Question 1

Twenty-five milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH before any NaOH is added?

Accepted Answer

The initial pH of the HCl solution can be calculated from the concentration of H+ ions in solution:
[H+] = 0.10 M HCl × (1 mol H+/1 mol HCl) = 0.10 M
pH = -log[H+] = -log(0.10) = 1.0

Question 2

What is the pH of a 1.0 M solution of trisodium phosphate? Ka1 = 7.1 × 10^-3, Ka2 = 6.3 × 10^-8, Ka3 = 4.2 × 10^-13

Accepted Answer

The answer of What is the pH of a 1.0...

Question 3

Determine the pH of the following solution. Initial concentrations are given. [NH₃] = 1.20 M, [KOH] = 0.320 M, Kb (NH₃) = 1.8 × 10^-5

Accepted Answer

The solution contains a strong base (KOH) and a weak base (NH3). The pH is primarily determined by the strong base, KOH. The concentration of OH- from KOH is 0.320 M, leading to a pOH of -log(0.320) = 0.49. Therefore, the pH is 14 - 0.49 = 13.51, approximately 13.5.

Question 4

25 ml of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH at the equivalence point? Ka for acetic acid = 1.8 × 10^-5.

Accepted Answer

The answer of 25 ml of 0.10 M acetic acid...

Question 5

What volume in ml of 0.05 M NaOH would have to be added to 50.0 ml of 0.10 M H₂SO₄ in order to affect complete neutralization of the acid?

Accepted Answer

H₂SO₄ is a diprotic acid, meaning it can donate two protons. The reaction is: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Therefore, 1 mole of H₂SO₄ reacts with 2 moles of NaOH. The moles of H₂SO₄ in 50 mL of 0.10 M solution is 0.005 moles. Thus, 0.01 moles of NaOH are needed. Using the formula Molarity = moles/volume, the volume of 0.05 M NaOH needed is 0.01 moles / 0.05 M = 0.2 L or 200 mL.

Question 6

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

Accepted Answer

The answer of What volume in mL of 2.0 M...

Question 7

What is the pH of a 1.0 M solution of Na₂SO₃? Ka₁ = 1.3 × 10^-2, Ka₂ = 6.2 × 10^-8 ?

Accepted Answer

The given compound is a long and complex molecule, but we can simplify it to NaX, where X represents the polyatomic anion ₂SO₃. The acid-base reaction for NaX in water is: NaX + H2O ⇌ HX + NaOH where HX is the acidic form of X. Since X has two acidic hydrogens, it can donate two protons to the solution. Therefore, we need to consider two equilibria: HX ⇌ H+ + X- X- ⇌ H+ + X₂SO₃- The corresponding acid dissociation constants are: Ka1 = [H+][X-]/[HX] = 1.3 × 10^-2 Ka2 = [H+][X₂SO₃-]/[X-] = 6.2 × 10^-8 At equilibrium, the concentration of NaOH is equal to that of HX, which is x. The concentration of NaX is 1.0 M (assuming complete dissolution). Therefore, at equilibrium: [NaX] = [HX] = x [NaOH] = x Using the Ka1 expression, we can write: Ka1 = x^2/(1.0 - x) = 1.3 × 10^-2 Solving for x gives: x = 2.95 × 10^-2 This is the concentration of H+ in the solution, which we can convert to pH: pH = -log[H+] = 10.53 This value is closer to answer choice D (10.4) than any other option.

Question 8

Calculate the pH of an aqueous solution that is 1.0 M Na₂CO₃. Ka₂ for H₂CO₃ is 4.7 × 10^-11.

Accepted Answer

The answer of Calculate the pH of an aqueous solution...

Question 9

What is the pH of a 0.30 M trisodium phosphate solution? [Ka for monohydrogen phosphate ion is 4.2 × 10^-13]

Accepted Answer

The answer of What is the pH of a 0.30...

Question 10

What is the pH of a 1.0 M solution of Na₃AsO₄?( Ka1 = 6 × 10^-3, Ka₂ = 1 × 10^-7, Ka₃ = 3 × 10^-12 )

Accepted Answer

This compound is a salt of a weak acid (H3AsO4) and a strong base (NaOH). The anion (AsO4 3-) is the conjugate base of the weak acid. The pH of a solution containing the salt of a weak acid and a strong base can be calculated using the following equation:
pH = pKa + log([base]/[acid])
where pKa is the dissociation constant of the weak acid (given in the problem), [base] is the concentration of the conjugate base (in this case, AsO4 3-), and [acid] is the concentration of the weak acid (H3AsO4).

Using the given equilibrium constants, we can calculate the pKa values for the three ionization steps of H3AsO4:
pKa1 = 3.22
pKa2 = 7.21
pKa3 = 11.50

The pH of the solution can be calculated using the equation above and assuming that the concentration of AsO4 3- is equal to the concentration of NaSAsO4:
pH = 8.21 + log([NaSAsO4]/[H3AsO4])
pH = 8.21 + log(1/1)
pH = 8.21

Therefore, the pH of the solution is 8.2, which is answer choice C.

Question 11

Determine the pH of the following solution. Initial concentrations are given. [NaCHO₂] = 0.815 M, [HBr] = 0.105 M, Ka (HCHO₂) = 1.8 × 10^-4

Accepted Answer

First, we need to write the equation for the dissociation of HCHO₂: HCHO₂ + H2O ⇌ CHO₂- + H3O+ The Ka value tells us that this is a weak acid, so we can use the following approximation: [H3O+] = √(Ka *[HCHO₂] / [CHO₂-]) Next, we need to calculate the concentration of CHO₂- using the initial concentration of NaCHO₂. Since NaCHO₂ is a salt of the weak acid, we can assume that it completely dissociates in water: NaCHO₂ → Na+ + CHO₂- [CHO₂-] = [NaCHO₂] = 0.815 M Now we can use the equation for the equilibrium constant of water to solve for [H3O+]: Kw = [H3O+][OH-] = 1.0 × 10^-14 at 25°C Assuming that [OH-] is very small compared to [H3O+], we can approximate this as: [H3O+] ≈ Kw / [OH-] Since we know that the solution is acidic (due to the presence of HBr), we can assume that [H3O+] > [OH-]. Therefore: [H3O+] ≈ Kw / [OH-] > Kw / [H3O+] [H3O+]^2 > Kw [H3O+] > √Kw = 1.0 × 10^-7 M Finally, we can use the equation for pH: pH = -log[H3O+] Substituting in our values, we get: [H3O+] = √(Ka *[HCHO₂] / [CHO₂-]) = √(1.8 × 10^-4 * 0.105 / 0.815) ≈ 1.41 × 10^-3 M pH = -log(1.41 × 10^-3) ≈ 4.27

Question 12

What is the pH at the equivalence point for the titration of 0.20 M nitrous acid by 0.20 M sodium hydroxide? [Ka for nitrous acid is 4.5 × 10^-4]

Accepted Answer

At the equivalence point, the moles of nitrous acid will be equal to the moles of sodium hydroxide, resulting in a solution containing only the conjugate base of nitrous acid (NO2-) and sodium ions. Since NO2- is a weak base, it will hydrolyze water to a small extent, generating OH- ions and increasing the pH of the solution. The pH can be calculated using the expression for the base dissociation constant (Kb) of NO2-:

Kb = Kw/Ka = (1.0 x 10^-14)/(4.5 x 10^-4) = 2.22 x 10^-11

At the equivalence point, [NO2-] = [OH-] = 0.1 M (assuming equal volumes of titrant added), so we can use the Kb expression to calculate the concentration of OH- ions produced by the hydrolysis of NO2-:

Kb = [OH-]^2/[NO2-] = (x^2)/(0.1 - x)

where x is the concentration of OH- ions formed. Since x will be small compared to 0.1, we can simplify the expression to:

Kb = x^2/0.1

0.1x 2.22 x 10^-11 = x^2

x = sqrt(2.22 x 10^-10) = 1.49 x 10^-5 M

Thus, the pOH at the equivalence point is:

pOH = -log(1.49 x 10^-5) = 4.83

And the pH at the equivalence point is:

pH = 14.00 - pOH = 9.17

Therefore, the best choice is C, 8.17 (which is the closest answer choice to the calculated pH of 9.17).

Question 13

25 mL of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH after 30 ml of NaOH have been added? Ka for acetic acid = 1.8 × 10^-5.

Accepted Answer

The answer of 25 mL of 0.10 M acetic acid...

Question 14

Twenty-five milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH when 30 ml of NaOH have been added?

Accepted Answer

The answer of Twenty-five milliliters of 0.10 M HCl is...

Question 15

If 25 mL of 0.20 M NaOH is added to 50 mL of 0.10 M HC₂H₃O₂ (Ka = 1.8 × 10^-5), what is the pH?

Accepted Answer

First, we need to write out the balanced equation for the reaction of HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0 (acetic acid) with NaOH:
HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0 + NaOH → NaCS1U1B12S1U1B0HS1U1B13S1U1B012S1U1B0 + H2O

Next, we need to determine how much of each species is present before the reaction occurs. We have 50 mL of 0.10 M HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0, which is equivalent to 0.005 mol HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0. We also have 25 mL of 0.20 M NaOH, which is equivalent to 0.005 mol NaOH.

Since NaOH is a strong base, we can assume that it will completely react with acetic acid to produce its conjugate base, sodium acetate. Therefore, after the reaction occurs, we will have 0.005 mol NaCS1U1B12S1U1B0HS1U1B13S1U1B012S1U1B0 and 0 mol HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0 remaining.

Now, we need to calculate the concentration of each species in the final solution, which we will use to determine the pH. We know the total volume of the solution is 75 mL (50 mL + 25 mL), so we can calculate the total moles of solute:
0.005 mol NaOH + 0.005 mol HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0 → 0.005 mol NaCS1U1B12S1U1B0HS1U1B13S1U1B012S1U1B0

The final volume is also 75 mL, so the concentration of the sodium acetate is:

C(NaCS1U1B12S1U1B0HS1U1B13S1U1B012S1U1B0) = 0.005 mol / 0.075 L = 0.067 M

Now, we can use the equilibrium expression for the dissociation of acetic acid to calculate the concentration of H+ ions in solution:

Ka = [H+][C2H3O2-] / [HC2H3O2]

We know the initial concentration of acetic acid and the final concentration of the conjugate base, so we can solve for [H+]:

Ka = 1.8 × 10S1U1P1-5S1S1P0 = x^2 / (0.005 - x)

Simplifying this expression and using the quadratic formula, we find that:

[H+] = 3.3 × 10S1U1P1-9S1S1P0 M

Finally, we can calculate the pH:

pH = -log[H+] = 8.78

Question 16

1.80 grams of an impure mixture containing sodium carbonate required 84.0 ml of 0.125 M H₂SO₄ for complete neutralization. What percent of the mixture is sodium carbonate?

Accepted Answer

The reaction between sodium carbonate and sulfuric acid is: Na2CO3 + H2SO4 → Na2SO4 + CO2 + H2O. The moles of H2SO4 used are 0.125 M * 0.084 L = 0.0105 moles. Since the reaction ratio is 1:1, moles of Na2CO3 are also 0.0105. The molar mass of Na2CO3 is 106 g/mol, so the mass of Na2CO3 is 0.0105 moles * 106 g/mol = 1.113 g. The percentage of Na2CO3 in the mixture is (1.113 g / 1.80 g) * 100% = 61.83%, approximately 62%.

Question 17

25 ml of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH before any NaOH is added? Ka for acetic acid = 1.8 × 10^-5.

Accepted Answer

The pH of a weak acid solution can be calculated using the formula: pH = -log(√(Ka × [HA])). For acetic acid, [HA] = 0.10 M and Ka = 1.8 × 10^-5. Plugging these values into the formula gives pH ≈ 2.9.

Question 18

25 mL of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH after 15 ml of NaOH have been added? Ka for acetic acid = 1.8 × 10^-5.

Accepted Answer

The answer of 25 mL of 0.10 M acetic acid...

Question 19

Twenty-five milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH after 15 ml of NaOH has been added?

Accepted Answer

The answer of Twenty-five milliliters of 0.10 M HCl is...

Quiz 17: Additional Aspects of Acid-Base Equilibria

Twenty-five milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH before any NaOH is added?

What is the pH of a 1.0 M solution of trisodium phosphate? Ka1 = 7.1 × 10^-3, Ka2 = 6.3 × 10^-8, Ka3 = 4.2 × 10^-13

Determine the pH of the following solution. Initial concentrations are given. [NH₃] = 1.20 M, [KOH] = 0.320 M, Kb (NH₃) = 1.8 × 10^-5

25 ml of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH at the equivalence point? Ka for acetic acid = 1.8 × 10^-5.

What volume in ml of 0.05 M NaOH would have to be added to 50.0 ml of 0.10 M H₂SO₄ in order to affect complete neutralization of the acid?

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

What is the pH of a 1.0 M solution of Na₂SO₃? Ka₁ = 1.3 × 10^-2, Ka₂ = 6.2 × 10^-8 ?

Calculate the pH of an aqueous solution that is 1.0 M Na₂CO₃. Ka₂ for H₂CO₃ is 4.7 × 10^-11.

What is the pH of a 0.30 M trisodium phosphate solution? [Ka for monohydrogen phosphate ion is 4.2 × 10^-13]

What is the pH of a 1.0 M solution of Na₃AsO₄?( Ka1 = 6 × 10^-3, Ka₂ = 1 × 10^-7, Ka₃ = 3 × 10^-12 )

Determine the pH of the following solution. Initial concentrations are given. [NaCHO₂] = 0.815 M, [HBr] = 0.105 M, Ka (HCHO₂) = 1.8 × 10^-4

What is the pH at the equivalence point for the titration of 0.20 M nitrous acid by 0.20 M sodium hydroxide? [Ka for nitrous acid is 4.5 × 10^-4]

25 mL of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH after 30 ml of NaOH have been added? Ka for acetic acid = 1.8 × 10^-5.

Twenty-five milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH when 30 ml of NaOH have been added?

If 25 mL of 0.20 M NaOH is added to 50 mL of 0.10 M HC₂H₃O₂ (Ka = 1.8 × 10^-5) , what is the pH?

1.80 grams of an impure mixture containing sodium carbonate required 84.0 ml of 0.125 M H₂SO₄ for complete neutralization. What percent of the mixture is sodium carbonate?

25 ml of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH before any NaOH is added? Ka for acetic acid = 1.8 × 10^-5.

25 mL of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH after 15 ml of NaOH have been added? Ka for acetic acid = 1.8 × 10^-5.

Twenty-five milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH after 15 ml of NaOH has been added?

Quiz 17: Additional Aspects of Acid-Base Equilibria

Twenty-five milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH before any NaOH is added?

What is the pH of a 1.0 M solution of trisodium phosphate? Ka1 = 7.1 × 10-3, Ka2 = 6.3 × 10-8, Ka3 = 4.2 × 10-13

Determine the pH of the following solution. Initial concentrations are given. [NH3] = 1.20 M, [KOH] = 0.320 M, Kb (NH3) = 1.8 × 10-5

25 ml of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH at the equivalence point? Ka for acetic acid = 1.8 × 10-5.

What volume in ml of 0.05 M NaOH would have to be added to 50.0 ml of 0.10 M H2SO4 in order to affect complete neutralization of the acid?

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

What is the pH of a 1.0 M solution of Na2SO3? Ka1 = 1.3 × 10-2, Ka2 = 6.2 × 10-8 ?

Calculate the pH of an aqueous solution that is 1.0 M Na2CO3. Ka2 for H2CO3 is 4.7 × 10-11.

What is the pH of a 0.30 M trisodium phosphate solution? [Ka for monohydrogen phosphate ion is 4.2 × 10-13]

What is the pH of a 1.0 M solution of Na3AsO4?( Ka1 = 6 × 10-3, Ka2 = 1 × 10-7, Ka3 = 3 × 10-12 )

Determine the pH of the following solution. Initial concentrations are given. [NaCHO2] = 0.815 M, [HBr] = 0.105 M, Ka (HCHO2) = 1.8 × 10-4

What is the pH at the equivalence point for the titration of 0.20 M nitrous acid by 0.20 M sodium hydroxide? [Ka for nitrous acid is 4.5 × 10-4]

25 mL of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH after 30 ml of NaOH have been added? Ka for acetic acid = 1.8 × 10-5.

Twenty-five milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH when 30 ml of NaOH have been added?

If 25 mL of 0.20 M NaOH is added to 50 mL of 0.10 M HC2H3O2 (Ka = 1.8 × 10-5) , what is the pH?

1.80 grams of an impure mixture containing sodium carbonate required 84.0 ml of 0.125 M H2SO4 for complete neutralization. What percent of the mixture is sodium carbonate?

25 ml of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH before any NaOH is added? Ka for acetic acid = 1.8 × 10-5.

25 mL of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH after 15 ml of NaOH have been added? Ka for acetic acid = 1.8 × 10-5.

Twenty-five milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH after 15 ml of NaOH has been added?

What is the pH of a 1.0 M solution of trisodium phosphate? Ka1 = 7.1 × 10^-3, Ka2 = 6.3 × 10^-8, Ka3 = 4.2 × 10^-13

Determine the pH of the following solution. Initial concentrations are given. [NH₃] = 1.20 M, [KOH] = 0.320 M, Kb (NH₃) = 1.8 × 10^-5

25 ml of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH at the equivalence point? Ka for acetic acid = 1.8 × 10^-5.

What volume in ml of 0.05 M NaOH would have to be added to 50.0 ml of 0.10 M H₂SO₄ in order to affect complete neutralization of the acid?

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

What is the pH of a 1.0 M solution of Na₂SO₃? Ka₁ = 1.3 × 10^-2, Ka₂ = 6.2 × 10^-8 ?

Calculate the pH of an aqueous solution that is 1.0 M Na₂CO₃. Ka₂ for H₂CO₃ is 4.7 × 10^-11.

What is the pH of a 0.30 M trisodium phosphate solution? [Ka for monohydrogen phosphate ion is 4.2 × 10^-13]

What is the pH of a 1.0 M solution of Na₃AsO₄?( Ka1 = 6 × 10^-3, Ka₂ = 1 × 10^-7, Ka₃ = 3 × 10^-12 )

Determine the pH of the following solution. Initial concentrations are given. [NaCHO₂] = 0.815 M, [HBr] = 0.105 M, Ka (HCHO₂) = 1.8 × 10^-4

What is the pH at the equivalence point for the titration of 0.20 M nitrous acid by 0.20 M sodium hydroxide? [Ka for nitrous acid is 4.5 × 10^-4]

25 mL of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH after 30 ml of NaOH have been added? Ka for acetic acid = 1.8 × 10^-5.

If 25 mL of 0.20 M NaOH is added to 50 mL of 0.10 M HC₂H₃O₂ (Ka = 1.8 × 10^-5) , what is the pH?

1.80 grams of an impure mixture containing sodium carbonate required 84.0 ml of 0.125 M H₂SO₄ for complete neutralization. What percent of the mixture is sodium carbonate?

25 ml of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH before any NaOH is added? Ka for acetic acid = 1.8 × 10^-5.

25 mL of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH after 15 ml of NaOH have been added? Ka for acetic acid = 1.8 × 10^-5.