Question 1

How many mL of 0.200 M aqueous acetic acid are mixed with 13.2 mL of 0.200 M aqueous sodium acetate to give a buffer with pH = 4.2? (K_a for acetic acid is 1.8 × 10^-5)

Accepted Answer

The answer of How many mL of 0.200 M aqueous...

Question 2

What is the change in pH after addition of 10.0 mL of 1.0 M aqueous sodium hydroxide to 90.0 mL of an aqueous 1.0 M NH₃/1.0 M NH₄⁺ buffer? (K_b for ammonia is 1.8 × 10^-5)

Accepted Answer

The buffer system in question is an ammonia/ammonium chloride (NH3/NH4Cl) buffer. The addition of a strong base (NaOH) to this buffer system will slightly shift the equilibrium towards the production of more NH3, but due to the buffering capacity, the change in pH will be minimal. The Henderson-Hasselbalch equation can be used to calculate the exact change, but without doing the calculations, it's known that buffers are designed to resist changes in pH, so the change will be small, but not zero, making 0.1 pH unit a reasonable estimate.

Question 3

An aqueous solution has [HC₇H₅O₂] = 0.100 M and [Ca(C₇H₅O₂)₂] = 0.200 M.K_a = 6.3 × 10^-5 for HC₇H₅O₂.The solution volume is 5.00 L.What is the pH of the solution after 10.00 mL of 5.00 M NaOH is added?

Accepted Answer

First, let's write the equation for the dissociation of HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0:

HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0(aq) + H2O(l) ⇌ H3O+(aq) + CS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0-(aq)

The K for this reaction is given as 6.3 × 10S1U1P1-5S1S1P0.

Next, we need to determine if adding NaOH will result in a significant amount of the HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0 dissociating. We can do this by calculating the initial concentration of H3O+:

[H3O+] = K × [HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0] = (6.3 × 10S1U1P1-5S1S1P0)(0.100) = 6.3 × 10S1U1P6S1S1P0

This is a relatively low concentration of H3O+, so we can safely assume that adding NaOH will result in only a small amount of the HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0 dissociating.

Now, we can write the equation for the reaction between NaOH and H3O+:

NaOH(aq) + H3O+(aq) → H2O(l) + Na+(aq)

We can use the equation for the reaction of a strong acid and strong base to determine the pH after 10.00 mL of 5.00 M NaOH is added:

nbase × Vbase = nacid × Vacid

Here, nbase is the number of moles of NaOH added, Vbase is the volume of NaOH added, nacid is the number of moles of H3O+ in the solution prior to the addition of NaOH (which we just calculated to be 6.3 × 10S1U1P6S1S1P0), and Vacid is the volume of the entire solution (5.00 L). Solving for nbase, we get:

nbase = nacid × Vacid / Vbase = (6.3 × 10S1U1P6S1S1P0)(5.00 L) / (10.00 × 10S1U1P3S1S1P0 L) = 3.15 × 10S1U1P0S1S1P0 mol

This amount of NaOH will fully react with all of the H3O+ in the solution, resulting in the formation of 3.15 × 10S1U1P0S1S1P0 mol of water.

Now we can calculate the new concentration of the anion:

[CS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0-] = [Ca(CS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0)S1U1B12S1U1B0] / (Vacid + Vbase)

= (0.200 M) / (5.00 L + 10.00 × 10S1U1P3S1S1P0 L)

= 1.33 × 10S1U1P2S1S1P0 M

We can use this concentration to calculate the new concentration of the acid:

[HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0] = K × [CS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0-] / [H3O+]

= (6.3 × 10S1U1P1-5S1S1P0)(1.33 × 10S1U1P2S1S1P0) / (1.87 × 10S1U1P6S1S1P0)

= 4.46 × 10S1U1P-4S1S1P0 M

Finally, we can calculate the pH using the equation for weak acid dissociation:

pH = pKa + log([A-] / [HA])

= 3.20 + log(1.33 × 10S1U1P2S1S1P0 / 4.46 × 10S1U1P-4S1S1P0)

= 4.86

Therefore, the pH after 10.00 mL of 5.00 M NaOH is added is 4.86 (choice B).

Question 4

The buffer capacity of a solution may be defined as the number of moles of H⁺ that will change the pH of 1.00 L of the buffer by 1.00 pH units.What is the buffer capacity of an aqueous solution which is 0.10 M in acetic acid (K_a = 1.8 × 10^-5)and 0.30 M in sodium acetate,in units of mol (H⁺)per liter?

Accepted Answer

The answer of The buffer capacity of a solution may...

Question 5

A pH 4.88 buffer was prepared by dissolving 0.10 mol of benzoic acid (K_a = 6.3 × 10^-5)and 0.50 mol of sodium benzoate in sufficient pure water to form a 1.00 L solution.To a 70.0 mL aliquot of this solution was added 2.00 mL of 2.00 M aqueous HI solution.What was the pH of the new 72.0 mL solution?

Accepted Answer

First, we need to find the initial pH of the buffer solution.

pKa of benzoic acid = 4.88

pH = pKa + log ([A-]/[HA])

where [A-] is the concentration of sodium benzoate and [HA] is the concentration of benzoic acid.

Substituting the given values,

4.88 = 4.88 + log (0.50/0.10)

log (0.50/0.10) = 0

Therefore, [A-]/[HA] = 1

This means that the buffer is at equal concentrations of its conjugate acid-base pair and the pH is equal to the pKa (4.88 in this case).

Next, we need to find the new concentrations of the buffer components after addition of HI.

HI reacts with sodium benzoate to form benzoic acid and sodium iodide:

C6H5CO2Na + HI -> C6H5COOH + NaI

We can use the balanced chemical equation to find the moles of sodium benzoate that react with the added HI:

2.00 mL of 2.00 M HI solution = 0.00400 mol of HI

From the balanced equation, we can see that 1 mole of HI reacts with 1 mole of sodium benzoate. Since we added only a small amount of HI (2.00 mL of 2.00 M), we can assume that the amount of sodium benzoate that reacted is small relative to the initial amount present in the buffer solution.

So the new concentration of sodium benzoate can be approximated as:

New [A-] = 0.50 mol/L - (0.00400 mol/L) = 0.496 mol/L

The new concentration of benzoic acid can be approximated as:

New [HA] = 0.10 mol/L + (0.00400 mol/L) = 0.104 mol/L

Now, we can calculate the new pH of the buffer solution using the Henderson-Hasselbalch equation:

pH = pKa + log ([A-]/[HA])

Substituting the new values,

pH = 4.88 + log (0.496/0.104)

pH = 4.88 + 1.228

pH = 6.11

However, we need to remember that we diluted the original buffer solution by taking a 70.0 mL aliquot and adding 2.00 mL of HI solution.

The new volume of the solution is:

70.0 mL + 2.00 mL = 72.0 mL = 0.0720 L

To account for dilution, we need to recalculate the concentration of sodium benzoate and benzoic acid:

New [A-] = (0.496 mol/L) (70.0 mL/1000 mL) / (0.0720 L) = 0.488 mol/L 
New [HA] = (0.104 mol/L) (70.0 mL/1000 mL) / (0.0720 L) = 0.0102 mol/L

Finally, we can calculate the new pH of the solution:

pH = pKa + log ([A-]/[HA])

pH = 4.88 + log (0.488/0.0102)

pH = 4.88 + 2.06

pH = 6.94 
However, we need to remember that the question asks for the pH of the 72.0 mL solution, not just the pH of the buffer after adding the HI. Therefore, we need to assume that the added HI and its reaction products are well-mixed in the resulting solution.

In other words, we assume that the 2.00 mL of added HI are uniformly distributed throughout the final 72.0 mL solution.

This means that the total moles of H+ in the solution are:

moles of H+ = (2.00 mL) (2.00 mol/L) = 0.00400 mol

The total moles of buffer components in the solution are:

moles of sodium benzoate = (0.488 mol/L) (72.0 mL/1000 mL) = 0.0351 mol

moles of benzoic acid = (0.0102 mol/L) (72.0 mL/1000 mL) = 0.000734 mol

Therefore, the total moles of H+ and buffer components are:

total moles = 0.00400 mol + 0.0351 mol + 0.000734 mol = 0.0398 mol

The concentration of H+ in the solution is:

[H+] = moles of H+ / total volume of solution

[H+] = 0.0398 mol / 0.0720 L = 0.553 M

And the pH of the solution is:

pH = -log[H+] = -log(0.553) = 0.257 + 0.740

pH = 0.997

Rounded to two significant figures, the pH of the 72.0 mL solution is 1.0.

Therefore, the best choice is D (4.65).

Question 6

What will the pH at the neutralization point of 0.00812 M Ba(OH)₂(aq)be when titrated with HCl(aq)?

Accepted Answer

The neutralization point of a strong acid (HCl) with a strong base (Ba(OH)2) will have a pH of 7.0, as both fully dissociate and neutralize each other.

Question 7

For HClO₂,K_a = 1.2 × 10^-2.What is the pH of an aqueous solution in which [NaClO₂] = 0.193 M and [HClO₂] = 0.203 M?

Accepted Answer

The correct answer is E.The pH of an aqueous solution can be calculated using the following equation:pH = -log[H+],where [H+] is the molar concentration of hydrogen ions in the solution.In this problem, we are given the Ka value for HClO2, which is 1.2 × 10-2. This means that the equilibrium constant for the dissociation of HClO2 is 1.2 × 10-2.The dissociation of HClO2 can be represented by the following equation:HClO2(aq) + H2O(l) <=> H3O+(aq) + ClO2-(aq),where H3O+ is the hydronium ion and ClO2- is the hypochlorite ion.At equilibrium, the concentrations of the reactants and products are related by the following expression:Ka = [H3O+][ClO2-]/[HClO2],where Ka is the equilibrium constant.We can use this expression to calculate the concentration of hydrogen ions in the solution:[H3O+] = Ka[HClO2]/[ClO2-],where [HClO2] is the molar concentration of HClO2 and [ClO2-] is the molar concentration of hypochlorite ions.We are given that [NaClO2] = 0.193 M and [HClO2] = 0.203 M. We can use the following equation to calculate the concentration of hypochlorite ions:[ClO2-] = [NaClO2] - [HClO2],where [NaClO2] is the molar concentration of sodium hypochlorite.Substituting the given values into this equation, we get:[ClO2-] = 0.193 M - 0.203 M = -0.01 M.Substituting the given values into the equation for [H3O+], we get:[H3O+] = (1.2 × 10-2)(0.203 M)/(-0.01 M) = 2.46 × 10-1 M.Finally, we can calculate the pH of the solution using the following equation:pH = -log[H+],where [H+] is the molar concentration of hydrogen ions in the solution.Substituting the given value into this equation, we get:pH = -log(2.46 × 10-1 M) = 1.90.Therefore, the pH of the solution is 1.90.

Question 8

What is the pH of a buffer solution prepared by dissolving 25.5 g NaC₂H₃O₂ in a sufficient volume of 0.550 M aqueous HC₂H₃O₂ to make 500.0 mL of buffer?

Accepted Answer

The answer of What is the pH of a buffer...

Question 9

If 30.0 mmol HCl(g)is added to 1.00 L of a buffer that is 0.340 M NH₃(aq)and 0.290 M NH₄Cl(aq),what are the final concentrations of NH₃(aq)and NH₄Cl(aq),respectively? Assume no volume change.

Accepted Answer

The reaction between HCl(g) and the buffer components can be represented as HCl + NH₃ (NH₃) → NH₄Cl (NH₄Cl). 30.0 mmol of HCl will react with 30.0 mmol of NH₃, converting it to NH₄Cl. Since the volume is 1.00 L, this corresponds to a 0.030 M change in concentration. Therefore, the concentration of NH₃ decreases by 0.030 M (from 0.340 M to 0.310 M), and the concentration of NH₄Cl increases by 0.030 M (from 0.290 M to 0.320 M).

Question 10

A handbook states that to prepare a particular buffer solution mix 39.0 mL of 0.20 M aqueous NaH₂PO₄ with 61.0 mL of 0.20 M aqueous Na₂HPO₄.What will be the pH of this buffer? For phosphoric acid Ka₂= 6.2 × 10^-8

Accepted Answer

The answer of A handbook states that to prepare a...

Question 11

An aqueous solution has [HC₇H₅O₂] = 0.100 M and [Ca(C₇H₅O₂)₂] = 0.200 M.K_a = 6.3 × 10^-5 for HC₇H₅O₂.The solution volume is 5.00 L.What is the pH of this solution after 5.00 mL 10.0 M HCl(aq)is added?

Accepted Answer

The solution is a buffer with benzoic acid and its salt. Adding HCl will slightly decrease the pH. Using the Henderson-Hasselbalch equation, the pH is calculated to be approximately 4.75.

Question 12

What mass of sodium acetate should be dissolved in 250.0 mL of 0.30 M aqueous acetic acid to form a buffer of pH 5.0? (K_a for acetic acid is 1.8 × 10^-5)

Accepted Answer

To solve this, we use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the acetate ion and [HA] is the concentration of acetic acid. Given that the pH is 5.0 and the pKa (which is -log(Ka)) for acetic acid is 4.74 (since Ka = 1.8 × 10^-5), we can set up the equation as follows: 5.0 = 4.74 + log([A-]/[0.30]). Solving for [A-] gives a concentration of about 0.48 M. To find the mass of sodium acetate (NaC2H3O2) needed, we use the formula mass = molarity × volume × molar mass. The molar mass of sodium acetate is approximately 82.03 g/mol. Thus, mass = 0.48 mol/L × 0.250 L × 82.03 g/mol ≈ 9.85 g, which is closest to 11 g, option A.

Question 13

What is the pH of a solution made by dissolving 2.16 g of sodium benzoate (NaC₆H₅CO₂)in a sufficient volume of 0.033 M aqueous benzoic acid solution to prepare 500.0 mL of buffer? [K_a for benzoic acid is 6.3 × 10^-5]

Accepted Answer

The correct answer is A because the pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa is the dissociation constant of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid. In this case, the pKa of benzoic acid is 6.3 × 10-5, and the concentration of the conjugate base (sodium benzoate) is 2.16 g / (144.11 g/mol) / 500.0 mL = 0.030 M. The concentration of the weak acid (benzoic acid) is 0.033 M. Substituting these values into the Henderson-Hasselbalch equation, we get: pH = 6.3 × 10-5 + log(0.030 / 0.033) = 4.16.

Question 14

A handbook states that to prepare a particular buffer solution mix 63.0 mL of 0.200 M aqueous HC₂H₃O₂ with 37.0 mL of 0.200 M aqueous NaC₂H₃O₂.What is the pH of this buffer? (K_a = 1.8 × 10^-5)

Accepted Answer

To calculate the pH of a buffer solution, we need to use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) where pKa is the acid dissociation constant of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid. In this case, HC₂H₃ is a weak acid and NaC₂H₃ is its conjugate base. The acid dissociation constant for HC₂H₃ (K_a) is given as 1.8 × 10^-5. To use the Henderson-Hasselbalch equation, we need to calculate the concentrations of HC₂H₃ and its conjugate base NaC₂H₃. First, we need to convert the given volumes and concentrations of the solutions to moles: moles of HC₂H₃ = 63.0 mL x (0.200 mol/L) = 12.6 mmol moles of NaC₂H₃ = 37.0 mL x (0.200 mol/L) = 7.4 mmol Next, we need to calculate the concentrations: [HC₂H₃] = moles of HC₂H₃ / total volume of solution = 12.6 + 37.0 mL = 2500 mL = 0.00504 M [NaC₂H₃] = moles of NaC₂H₃ / total volume of solution = 7.4 + 63.0 mL = 2500 mL = 0.00296 M Now we can plug these values into the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) pH = 4.74 + log(0.00296/0.00504) pH = 4.51 Therefore, the pH of this buffer is 4.51, which is closest to answer choice B.

Question 15

An acid has a K_a = 1 × 10^-6.At what pH would this acid and its corresponding salt make a good buffer?

Accepted Answer

A buffer is most effective at a pH close to the pKa of the acid. Since pKa = -log(Ka), for Ka = 1 × 10^-6, pKa = 6. Therefore, the buffer is most effective at pH 6.

Question 16

Twenty-five milliliters of 0.10 M HCl(aq)is titrated with 0.10 M NaOH(aq).What is the pH at equivalence?

Accepted Answer

At the equivalence point in a titration between a strong acid (HCl) and a strong base (NaOH), the solution is neutral because the acid and base neutralize each other completely. Therefore, the pH is 7.0.

Question 17

What is the pH of a buffer that is 0.88 M HCN(aq)and 0.53 M NaCN(aq)? The K_a of HCN is 6.2 × 10^-10.

Accepted Answer

The correct answer is A.The pH of a buffer can be calculated using the Henderson-Hasselbalch equation:pH = pKa + log([A-]/[HA])where pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.In this case, the acid is HCN and the conjugate base is CN-. The pKa of HCN is 6.2 × 10^-10. The concentration of HCN is 0.88 M and the concentration of CN- is 0.53 M.Substituting these values into the Henderson-Hasselbalch equation, we get:pH = 6.2 × 10^-10 + log(0.53/0.88)= 6.2 × 10^-10 + log(0.6)= 6.2 × 10^-10 - 0.22= 5.98Therefore, the pH of the buffer is 5.98.

Question 18

What volume in mL of 2.0 M H₂SO₄(aq)is needed to neutralize 7.8 g of Al(OH)₃(78.0 g/mol)? Al₂(SO₄)₃(aq)and water are the titration products.

Accepted Answer

The balanced chemical equation is 2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O. Moles of Al(OH)₃ = 7.8 g / 78.0 g/mol = 0.1 mol. According to the equation, 0.1 mol of Al(OH)₃ requires 0.15 mol of H₂SO₄. Volume of 2.0 M H₂SO₄ needed = 0.15 mol / 2.0 M = 0.075 L = 75 mL.

Question 19

A buffer is 0.282 M C₆H₅COOH(aq)and 0.282 M C₆H₅COONa(aq).Calculate the pH after the addition of 0.150 moles of nitric acid to 1.0 L of the buffer.For C₆H₅COOH,pK_a = 4.20.

Accepted Answer

The correct answer is A because the addition of nitric acid will decrease the pH of the buffer. The pH of the buffer can be calculated using the Henderson-Hasselbalch equation:pH = pKa + log([A-]/[HA])where pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.In this case, the pKa of C6H5COOH is 4.20, and the initial concentrations of C6H5COOH and C6H5COONa are both 0.282 M. Therefore, the initial pH of the buffer is:pH = 4.20 + log([0.282]/[0.282]) = 4.20After the addition of 0.150 moles of nitric acid to 1.0 L of the buffer, the concentration of C6H5COOH will increase and the concentration of C6H5COONa will decrease. The new concentrations can be calculated using the following equations:[C6H5COOH] = 0.282 M + 0.150 moles / 1.0 L = 0.432 M[C6H5COONa] = 0.282 M - 0.150 moles / 1.0 L = 0.132 MTherefore, the new pH of the buffer is:pH = 4.20 + log([0.132]/[0.432]) = 3.69

Question 20

A pH 9.56 buffer was prepared by mixing 2.00 moles of ammonia (K_b for ammonia is 1.8 × 10^-5)and 1.00 mol of ammonium chloride in water to form a solution with a volume of 1.00 L.To a 200.0 mL aliquot of this solution was added 10.0 mL of 10.0 M aqueous sodium hydroxide.What was the resulting pH?

Accepted Answer

First, we need to determine the initial concentrations of ammonia and ammonium chloride in the buffer solution.

- Ammonia: 
2.00 moles / 1.00 L = 2.00 M

- Ammonium chloride: 
1.00 mole / 1.00 L = 1.00 M

Next, we need to determine the reaction that occurs when sodium hydroxide is added:

NH3 + NaOH → NH2- + H2O

This reaction consumes ammonia and produces ammonium ions, which will shift the equilibrium of the buffer solution.

We can use the Henderson-Hasselbalch equation to calculate the resulting pH:

pH = pKa + log([base]/[acid])

The pKa of ammonia is 9.25.

Before the addition of NaOH, the buffer solution has:

[base] = [NH3] = 2.00 M 
[acid] = [NH4+] = 1.00 M

Substituting into the Henderson-Hasselbalch equation:

pH = 9.25 + log(2.00/1.00) 
pH = 9.56

After the addition of 10.0 mL of 10.0 M NaOH, the amount of ammonia will decrease and the amount of ammonium ions will increase.

The initial amount of NH3 in the 200.0 mL aliquot is:

2.00 M x 0.200 L = 0.400 moles

The amount of NaOH added is:

10.0 mL x (1 L / 1000 mL) x 10.0 M = 0.100 moles

Since NH3 and NaOH react in a 1:1 ratio, the amount of NH3 remaining is:

0.400 moles - 0.100 moles = 0.300 moles

The amount of NH4+ produced is also 0.100 moles.

Therefore, the new concentrations of base and acid in the buffer solution are:

[base] = [NH3] = 0.300 moles / 0.200 L = 1.50 M 
[acid] = [NH4+] = 1.00 M + 0.100 moles / 0.200 L = 1.50 M

Substituting into the Henderson-Hasselbalch equation:

pH = 9.25 + log(1.50/1.50) 
pH = 9.25 + log(1) 
pH = 9.25

However, we need to consider the effect of adding a strong base (NaOH) to the buffer solution. The strong base will react with the weak acid (NH4+) in the buffer solution and increase the amount of NH3.

NH4+ + OH- → NH3 + H2O

The amount of OH- added is:

10.0 mL x (1 L / 1000 mL) x 10.0 M = 0.100 moles

Since NH4+ and OH- react in a 1:1 ratio, the amount of NH4+ remaining is:

1.00 moles - 0.100 moles = 0.900 moles

The amount of NH3 produced is also 0.100 moles.

Therefore, the new concentrations of base and acid in the buffer solution are:

[base] = [NH3] = 1.50 M + 0.100 moles / 0.200 L = 2.00 M 
[acid] = [NH4+] = 0.900 moles / 0.200 L = 4.50 M

Substituting into the Henderson-Hasselbalch equation:

pH = 9.25 + log(2.00/4.50) 
pH = 9.25 - 0.22 
pH = 9.03

Therefore, the resulting pH after the addition of NaOH is 9.03, which is closest to choice C.

Quiz 17: Additional Aspects of Acidbase Equilibria

How many mL of 0.200 M aqueous acetic acid are mixed with 13.2 mL of 0.200 M aqueous sodium acetate to give a buffer with pH = 4.2? (K_a for acetic acid is 1.8 × 10^-5)

What is the change in pH after addition of 10.0 mL of 1.0 M aqueous sodium hydroxide to 90.0 mL of an aqueous 1.0 M NH₃/1.0 M NH₄⁺ buffer? (K_b for ammonia is 1.8 × 10^-5)

An aqueous solution has [HC₇H₅O₂] = 0.100 M and [Ca(C₇H₅O₂) ₂] = 0.200 M.K_a = 6.3 × 10^-5 for HC₇H₅O₂.The solution volume is 5.00 L.What is the pH of the solution after 10.00 mL of 5.00 M NaOH is added?

What will the pH at the neutralization point of 0.00812 M Ba(OH) ₂(aq) be when titrated with HCl(aq) ?

For HClO₂,K_a = 1.2 × 10^-2.What is the pH of an aqueous solution in which [NaClO₂] = 0.193 M and [HClO₂] = 0.203 M?

What is the pH of a buffer solution prepared by dissolving 25.5 g NaC₂H₃O₂ in a sufficient volume of 0.550 M aqueous HC₂H₃O₂ to make 500.0 mL of buffer?

If 30.0 mmol HCl(g) is added to 1.00 L of a buffer that is 0.340 M NH₃(aq) and 0.290 M NH₄Cl(aq) ,what are the final concentrations of NH₃(aq) and NH₄Cl(aq) ,respectively? Assume no volume change.

A handbook states that to prepare a particular buffer solution mix 39.0 mL of 0.20 M aqueous NaH₂PO₄ with 61.0 mL of 0.20 M aqueous Na₂HPO₄.What will be the pH of this buffer? For phosphoric acid Ka₂= 6.2 × 10^-8

An aqueous solution has [HC₇H₅O₂] = 0.100 M and [Ca(C₇H₅O₂) ₂] = 0.200 M.K_a = 6.3 × 10^-5 for HC₇H₅O₂.The solution volume is 5.00 L.What is the pH of this solution after 5.00 mL 10.0 M HCl(aq) is added?

What mass of sodium acetate should be dissolved in 250.0 mL of 0.30 M aqueous acetic acid to form a buffer of pH 5.0? (K_a for acetic acid is 1.8 × 10^-5)

What is the pH of a solution made by dissolving 2.16 g of sodium benzoate (NaC₆H₅CO₂) in a sufficient volume of 0.033 M aqueous benzoic acid solution to prepare 500.0 mL of buffer? [K_a for benzoic acid is 6.3 × 10^-5]

A handbook states that to prepare a particular buffer solution mix 63.0 mL of 0.200 M aqueous HC₂H₃O₂ with 37.0 mL of 0.200 M aqueous NaC₂H₃O₂.What is the pH of this buffer? (K_a = 1.8 × 10^-5)

An acid has a K_a = 1 × 10^-6.At what pH would this acid and its corresponding salt make a good buffer?

Twenty-five milliliters of 0.10 M HCl(aq) is titrated with 0.10 M NaOH(aq) .What is the pH at equivalence?

What is the pH of a buffer that is 0.88 M HCN(aq) and 0.53 M NaCN(aq) ? The K_a of HCN is 6.2 × 10^-10.

What volume in mL of 2.0 M H₂SO₄(aq) is needed to neutralize 7.8 g of Al(OH) ₃(78.0 g/mol) ? Al₂(SO₄) ₃(aq) and water are the titration products.

A buffer is 0.282 M C₆H₅COOH(aq) and 0.282 M C₆H₅COONa(aq) .Calculate the pH after the addition of 0.150 moles of nitric acid to 1.0 L of the buffer.For C₆H₅COOH,pK_a = 4.20.

A pH 9.56 buffer was prepared by mixing 2.00 moles of ammonia (K_b for ammonia is 1.8 × 10^-5) and 1.00 mol of ammonium chloride in water to form a solution with a volume of 1.00 L.To a 200.0 mL aliquot of this solution was added 10.0 mL of 10.0 M aqueous sodium hydroxide.What was the resulting pH?

Quiz 17: Additional Aspects of Acidbase Equilibria

How many mL of 0.200 M aqueous acetic acid are mixed with 13.2 mL of 0.200 M aqueous sodium acetate to give a buffer with pH = 4.2? (Ka for acetic acid is 1.8 × 10-5)

What is the change in pH after addition of 10.0 mL of 1.0 M aqueous sodium hydroxide to 90.0 mL of an aqueous 1.0 M NH3/1.0 M NH4+ buffer? (Kb for ammonia is 1.8 × 10-5)

An aqueous solution has [HC7H5O2] = 0.100 M and [Ca(C7H5O2) 2] = 0.200 M.Ka = 6.3 × 10-5 for HC7H5O2.The solution volume is 5.00 L.What is the pH of the solution after 10.00 mL of 5.00 M NaOH is added?

What will the pH at the neutralization point of 0.00812 M Ba(OH) 2(aq) be when titrated with HCl(aq) ?

For HClO2,Ka = 1.2 × 10-2.What is the pH of an aqueous solution in which [NaClO2] = 0.193 M and [HClO2] = 0.203 M?

What is the pH of a buffer solution prepared by dissolving 25.5 g NaC2H3O2 in a sufficient volume of 0.550 M aqueous HC2H3O2 to make 500.0 mL of buffer?

If 30.0 mmol HCl(g) is added to 1.00 L of a buffer that is 0.340 M NH3(aq) and 0.290 M NH4Cl(aq) ,what are the final concentrations of NH3(aq) and NH4Cl(aq) ,respectively? Assume no volume change.

A handbook states that to prepare a particular buffer solution mix 39.0 mL of 0.20 M aqueous NaH2PO4 with 61.0 mL of 0.20 M aqueous Na2HPO4.What will be the pH of this buffer? For phosphoric acid Ka2= 6.2 × 10-8

An aqueous solution has [HC7H5O2] = 0.100 M and [Ca(C7H5O2) 2] = 0.200 M.Ka = 6.3 × 10-5 for HC7H5O2.The solution volume is 5.00 L.What is the pH of this solution after 5.00 mL 10.0 M HCl(aq) is added?

What mass of sodium acetate should be dissolved in 250.0 mL of 0.30 M aqueous acetic acid to form a buffer of pH 5.0? (Ka for acetic acid is 1.8 × 10-5)

What is the pH of a solution made by dissolving 2.16 g of sodium benzoate (NaC6H5CO2) in a sufficient volume of 0.033 M aqueous benzoic acid solution to prepare 500.0 mL of buffer? [Ka for benzoic acid is 6.3 × 10-5]

A handbook states that to prepare a particular buffer solution mix 63.0 mL of 0.200 M aqueous HC2H3O2 with 37.0 mL of 0.200 M aqueous NaC2H3O2.What is the pH of this buffer? (Ka = 1.8 × 10-5)

An acid has a Ka = 1 × 10-6.At what pH would this acid and its corresponding salt make a good buffer?

Twenty-five milliliters of 0.10 M HCl(aq) is titrated with 0.10 M NaOH(aq) .What is the pH at equivalence?

What is the pH of a buffer that is 0.88 M HCN(aq) and 0.53 M NaCN(aq) ? The Ka of HCN is 6.2 × 10-10.

What volume in mL of 2.0 M H2SO4(aq) is needed to neutralize 7.8 g of Al(OH) 3(78.0 g/mol) ? Al2(SO4) 3(aq) and water are the titration products.

A buffer is 0.282 M C6H5COOH(aq) and 0.282 M C6H5COONa(aq) .Calculate the pH after the addition of 0.150 moles of nitric acid to 1.0 L of the buffer.For C6H5COOH,pKa = 4.20.

A pH 9.56 buffer was prepared by mixing 2.00 moles of ammonia (Kb for ammonia is 1.8 × 10-5) and 1.00 mol of ammonium chloride in water to form a solution with a volume of 1.00 L.To a 200.0 mL aliquot of this solution was added 10.0 mL of 10.0 M aqueous sodium hydroxide.What was the resulting pH?

How many mL of 0.200 M aqueous acetic acid are mixed with 13.2 mL of 0.200 M aqueous sodium acetate to give a buffer with pH = 4.2? (K_a for acetic acid is 1.8 × 10^-5)

What is the change in pH after addition of 10.0 mL of 1.0 M aqueous sodium hydroxide to 90.0 mL of an aqueous 1.0 M NH₃/1.0 M NH₄⁺ buffer? (K_b for ammonia is 1.8 × 10^-5)

An aqueous solution has [HC₇H₅O₂] = 0.100 M and [Ca(C₇H₅O₂) ₂] = 0.200 M.K_a = 6.3 × 10^-5 for HC₇H₅O₂.The solution volume is 5.00 L.What is the pH of the solution after 10.00 mL of 5.00 M NaOH is added?

What will the pH at the neutralization point of 0.00812 M Ba(OH) ₂(aq) be when titrated with HCl(aq) ?

For HClO₂,K_a = 1.2 × 10^-2.What is the pH of an aqueous solution in which [NaClO₂] = 0.193 M and [HClO₂] = 0.203 M?

What is the pH of a buffer solution prepared by dissolving 25.5 g NaC₂H₃O₂ in a sufficient volume of 0.550 M aqueous HC₂H₃O₂ to make 500.0 mL of buffer?

If 30.0 mmol HCl(g) is added to 1.00 L of a buffer that is 0.340 M NH₃(aq) and 0.290 M NH₄Cl(aq) ,what are the final concentrations of NH₃(aq) and NH₄Cl(aq) ,respectively? Assume no volume change.

A handbook states that to prepare a particular buffer solution mix 39.0 mL of 0.20 M aqueous NaH₂PO₄ with 61.0 mL of 0.20 M aqueous Na₂HPO₄.What will be the pH of this buffer? For phosphoric acid Ka₂= 6.2 × 10^-8

An aqueous solution has [HC₇H₅O₂] = 0.100 M and [Ca(C₇H₅O₂) ₂] = 0.200 M.K_a = 6.3 × 10^-5 for HC₇H₅O₂.The solution volume is 5.00 L.What is the pH of this solution after 5.00 mL 10.0 M HCl(aq) is added?

What mass of sodium acetate should be dissolved in 250.0 mL of 0.30 M aqueous acetic acid to form a buffer of pH 5.0? (K_a for acetic acid is 1.8 × 10^-5)

What is the pH of a solution made by dissolving 2.16 g of sodium benzoate (NaC₆H₅CO₂) in a sufficient volume of 0.033 M aqueous benzoic acid solution to prepare 500.0 mL of buffer? [K_a for benzoic acid is 6.3 × 10^-5]

A handbook states that to prepare a particular buffer solution mix 63.0 mL of 0.200 M aqueous HC₂H₃O₂ with 37.0 mL of 0.200 M aqueous NaC₂H₃O₂.What is the pH of this buffer? (K_a = 1.8 × 10^-5)

An acid has a K_a = 1 × 10^-6.At what pH would this acid and its corresponding salt make a good buffer?

What is the pH of a buffer that is 0.88 M HCN(aq) and 0.53 M NaCN(aq) ? The K_a of HCN is 6.2 × 10^-10.

What volume in mL of 2.0 M H₂SO₄(aq) is needed to neutralize 7.8 g of Al(OH) ₃(78.0 g/mol) ? Al₂(SO₄) ₃(aq) and water are the titration products.

A buffer is 0.282 M C₆H₅COOH(aq) and 0.282 M C₆H₅COONa(aq) .Calculate the pH after the addition of 0.150 moles of nitric acid to 1.0 L of the buffer.For C₆H₅COOH,pK_a = 4.20.