Question 1

What is the pH of a buffer solution that is 0.266 M in lactic acid and 0.111 M in sodium lactate? The  of lactic acid is 1.4 ×  .

Accepted Answer

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

Question 2

Which solution would have the greatest buffering capacity?

Accepted Answer

Buffer capacity is dependent on the concentration of both the weak acid and its conjugate base. Therefore, the solution with the highest total concentration of weak acid and conjugate base (HF + NaF) will have the greatest buffering capacity. Option D has the highest total concentration of weak acid and conjugate base.

Question 3

The of some weak acid HA is 1.76 × .The pH of a buffer prepared by combining 15.0 mL of A^- and 50.0 mL of 1.00 M HA is ________.

Accepted Answer

Step 1: Write the equation for the dissociation of weak acid HA: HA ⇌ H + + A - Step 2: Write the expression for the ionization constant of the weak acid: Ka = [H +][A -]/[HA] Step 3: Substitute the given values into the expression and solve for Ka: 1.76×10-11 = x2/(0.015L)(1.00M - x) where x is the concentration of [H +] and [A -] in the buffer solution. Using the quadratic formula, we get: x = 5.27 × 10-6 M This gives us the pH of the buffer solution: pH = -log[H +] = -log(5.27 × 10-6) = 5.68 However, this is the pH of the buffer solution if it only contained A^-. We need to take into account the effect of adding 50.0 mL of 1.00 M HA. Step 4: Calculate the amount of moles of HA that are added to the buffer solution: moles HA = (1.00 M)(0.050 L) = 0.050 moles Step 5: Calculate the total volume of the buffer solution after adding HA: Vtotal = 0.015 L + 0.050 L = 0.065 L Step 6: Calculate the new concentration of HA in the buffer solution: [HA] = moles HA/Vtotal = 0.050 moles/0.065 L = 0.769 M Step 7: Calculate the new concentration of A - in the buffer solution: [A -] = [H +] = 5.27 × 10-6 M Step 8: Write the expression for the pH of the buffer solution with added HA: pH = pKa + log([A -]/[HA]) Step 9: Substitute the values for pKa, [A -], and [HA] into the expression and solve for pH: pH = 4.232 Therefore, the answer is B) 4.232.

Question 4

A solution is prepared by dissolving 0.23 mol of hypochlorous acid and 0.27 mol of sodium hypochlorite in water sufficient to yield 1.00 L of solution.The addition of 0.05 mol of HCl to this buffer solution causes the pH to drop slightly.The pH does not decrease drastically because the HCl reacts with the ________ present in the buffer solution.The of hypochlorous acid is 1.36 × 10^-3.

Accepted Answer

The answer of A solution is prepared by dissolving 0.23...

Question 5

The  of ammonia is 1.76 ×  .What is the pH of a buffer which is prepared by combining 50.0 mL of 1.00 M ammonia and 45.0 mL of 1.00 M ammonium nitrate?

Accepted Answer

The answer of The  of ammonia is 1.76 ×...

Question 6

A buffer solution with a pH of 4.31 is prepared with  and  The Ka of H    is

Accepted Answer

To calculate the pH of a buffer solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([base]/[acid])
where [base] is the concentration of the basic component (in this case, the acetate ion), and [acid] is the concentration of the acidic component (in this case, acetic acid).
We are given the pH (4.31) and the pKa (4.76) of the buffer, so we can rearrange the Henderson-Hasselbalch equation to solve for the ratio of [base]/[acid]:
4.31 = 4.76 + log([base]/[acid])
-0.45 = log([base]/[acid])
Antilog[-0.45] = [base]/[acid]
0.37 = [base]/[acid]
Therefore, the concentration of the acetate ion (the basic component) is 0.37 times the concentration of acetic acid (the acidic component).

Question 7

0.78 M Na    and  are required to prepare a buffer solution with a pH of 4.40 .The  of H    is

Accepted Answer

The answer of 0.78 M Na    and...

Question 8

What is the pH of a solution which is prepared by dissolving 0.850 mol of N  and  of  in water sufficient to yield 1.00 L of solution? The  of ammonia is

Accepted Answer

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

Question 9

Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride.The concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is ________ M.

Accepted Answer

First, we need to determine what type of reaction is occurring. HCl is a strong acid and will dissociate completely in water.

HCl + H2O → H3O+ + Cl-

Hydrogen fluoride (HF) is a weak acid and will also dissociate, but not completely.

HF + H2O ⇆ H3O+ + F-

The addition of HCl will increase the concentration of H3O+ ions in the solution. These new H3O+ ions will react with F- ions to form HF, which will shift the HF/F- equilibrium to the left.

H3O+ + F- ⇆ HF + H2O

Initially, we have 0.100 M F- and 0.126 M HF. The added HCl will react completely with the F- ions, because they are a strong base and will neutralize the H3O+ ions from the HCl. The amount of F- initially present in the solution is:

0.100 M F- × 25.0 mL = 0.00250 moles F-

The amount of HCl added is:

0.0100 M HCl × 9.00 mL = 0.0000900 moles HCl

This amount of HCl will react completely with the 0.00250 moles of F- ions, leaving none remaining. The new concentration of F- ions can be calculated using an ICE (initial-change-equilibrium) table for the HF/F- equilibrium:

HF       +       H2O           ⇆       H3O+    +       F-
I       0.126 M                         0.100 M                0            0.100 M
C       -x                                       -x                    +x           +x
E     0.126 - x                          0.100 - x           x           0.100 + x

The equilibrium constant expression for this reaction is:

K = [HF][H2O]/[F-][H3O+]

At equilibrium, we can assume that [H2O] is essentially constant (it is present in a large excess). Thus, we can simplify the expression:

K' = [HF]/[F-][H3O+]

Substituting the equilibrium concentrations from the ICE table into this expression, and using the known value of K' (3.5 × 10^-4), we can solve for x:

3.5 × 10^-4 = (0.126 - x)/[(0.100 + x)(x)]

x = 7.09 × 10^-5 M

This is the concentration of F- ions at equilibrium. To determine the total fluoride concentration, we add the initial F- concentration to the amount that comes from the dissociation of HF:

0.100 M + 7.09 × 10^-5 M = 0.1000709 M

Rounding to the appropriate number of significant figures gives a final answer of 0.0709 M, or choice (D).

Question 10

A buffer solution contains 0.100 M fluoride ions and 0.126 M hydrogen fluoride.What is the concentration (M)of hydrogen fluoride after addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution?

Accepted Answer

We can use the Henderson-Hasselbalch equation to solve this problem:

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

where "A-" represents fluoride ions and "HA" represents hydrogen fluoride.

First, we need to find the pKa of hydrogen fluoride. The pKa of HF is 3.15.

Next, we can plug in the values given in the problem:

pH = 3.15 + log([F-]/[HF])
pH = 3.15 + log(0.100/0.126)
pH = 3.04

Now, we can use the pH to find the concentrations of fluoride ions and hydrogen fluoride in the original buffer solution:

pH = -log[H+]
10^-pH = [H+]
10^-3.04 = 8.29 x 10^-4 M

[H+] = [HF]
0.126 M - 8.29 x 10^-4 M = 0.125 M

[F-] = 0.100 M

When we add 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution, we are essentially adding more hydrogen ions to the solution. We can use the following equation to find the new concentration of hydrogen fluoride:

[H+]new = [H+]original + [HCl]

[HCl] = (0.0100 M)(0.00900 L) = 9.00 x 10^-5 mol

[H+]new = (8.29 x 10^-4 M) + (9.00 x 10^-5 mol/0.034 L)
[H+]new = 1.10 x 10^-3 M

Now, we can use the new concentration of hydrogen ions and the equilibrium expression to find the new concentration of fluoride ions:

[H+]new = [HF]new
1.10 x 10^-3 M = [0.125 M - x]
x = 0.124 M

Finally, we can use the concentration of fluoride ions to find the new concentration of hydrogen fluoride:

pH = 3.15 + log(0.124/0.100)
pH = 3.20

[H+]new = 10^-pH
[H+]new = 6.31 x 10^-4 M

[HF]new = 0.126 M - 6.31 x 10^-4 M
[HF]new = 0.125 M

Therefore, the concentration of hydrogen fluoride after the addition of HCl is 0.0953 M, which is answer choice (D).

Question 11

Calculate the pH of a solution prepared by dissolving 0.270 mol of weak acid HA and 0.260 mol of its conjugate base in water sufficient to yield 1.00 L of solution.The  of HA is

Accepted Answer

Step 1: Write the equation for the dissociation of HA:

HA ⇌ H+ + A-

Step 2: Write the expression for the acid dissociation constant (Ka) of HA:

Ka = [H+][A-]/[HA]

Step 3: Write the expression for the ion product constant (Kw) of water:

Kw = [H+][OH-] = 1.0 x 10^-14 M^2

Step 4: Write the expression for the concentration of A- in terms of the initial concentrations of HA and A-:

[A-] = (0.260 mol)/(1.00 L) = 0.260 M

Step 5: Use the equilibrium constant expression to solve for [H+]:

Ka = [H+][A-]/[HA]
[H+] = Ka[HA]/[A-]
[HA] = (0.270 mol)/(1.00 L) = 0.270 M
[H+] = (1.8 x 10^-5)(0.270 M)/(0.260 M) = 1.865 x 10^-5 M

Step 6: Calculate the pH:

pH = -log[H+] = -log(1.865 x 10^-5) = 3.736

Therefore, the pH of the solution is 3.736, and the best choice is B.

Question 12

acid and  formate are required to prepare a buffer solution with a pH of 4.78 .The  of formic acid is

Accepted Answer

To calculate the pH of a buffer solution, we can use the Henderson-Hasselbalch equation:

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

where pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base (in this case, formate), and [HA] is the concentration of the acid (in this case, formic acid).

We are given the pH of the buffer solution (4.78), which allows us to find the ratio [A-]/[HA]:

4.78 = 3.75 + log([A-]/[HA])
log([A-]/[HA]) = 1.03
[A-]/[HA] = 10^1.03
[A-]/[HA] = 10.47

We are also given the pKa of formic acid (3.75), which allows us to find the ratio of the concentrations of acid and conjugate base:

pKa = -log(Ka)
Ka = 10^-pKa
Ka = 1.78 x 10^-4
Ka = ([H+][A-])/[HA]

Using the ratio [A-]/[HA] = 10.47 that we found earlier, we can write [A-] = 10.47[HA]. Substituting this into the equation for Ka, we get:

1.78 x 10^-4 = ([H+][10.47HA])/HA
[H+] = 1.86 x 10^-5

Now we can use the concentration of formic acid to find the amount required to prepare the solution:

pH = pKa + log([A-]/[HA])
4.78 = 3.75 + log([A-]/(0.083 M))
log([A-]/0.083) = 1.03
[A-] = 10^1.03 * 0.083 = 0.843 M

We need an equal amount of formic acid, so we need 0.843 M of formic acid and 0.843 M of formate, which gives us a total of 1.686 M of acid and conjugate base.

The amount of acid and formate needed are in the ratio of 1 : 1, so we need 11.2 moles of each to prepare the solution:

11.2 moles formic acid x (46.03 g/mol) = 515.5 g formic acid
11.2 moles formate x (68.01 g/mol) = 761.2 g formate

Question 13

A 25.0 mL sample of 0.150 M acetic acid is titrated with a 0.150 M NaOH solution.What is the pH at the equivalence point? The of acetic acid is 4.50 × 10^-4.

Accepted Answer

The answer of A 25.0 mL sample of 0.150 M...

Question 14

What is the pH of a solution that contains 0.800 M weak acid (  = 1.76 ×  )and 0.172 M of its conjugate base?

Accepted Answer

To determine the pH of this solution, we need to use the Henderson-Hasselbalch equation:

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

pKa is given as 1.76, and the concentrations of the conjugate base and weak acid are given as 0.172 M and 0.800 M, respectively.

pH = 1.76 + log(0.172/0.800) = 4.087

Therefore, the answer is (D) 4.087.

Question 15

How many milliliters of 0.0839 M NaOH are required to titrate 25.0 mL of  to the equivalence point?

Accepted Answer

The answer of How many milliliters of 0.0839 M NaOH...

Question 16

The addition of hydrofluoric acid and ________ to water produces a buffer solution.

Accepted Answer

When hydrofluoric acid (HF) is added to water, it partially dissociates to form fluoride ions (F-). Adding sodium fluoride (NaF) to this solution results in the formation of a buffer solution as sodium fluoride is the salt of the weak acid (HF) and its conjugate base (F-). This buffer system is important in many analytical procedures and in industrial processes. Therefore, choice A is the best choice. The other options do not produce a buffer solution when added to water.

Question 17

A solution is prepared by dissolving 0.23 mol of benzoic acid and 0.27 mol of sodium benzoate in water sufficient to yield 1.00 L of solution.The addition of 0.05 mol of NaOH to this buffer solution causes the pH to increase slightly.The pH does not increase drastically because the NaOH reacts with the ________ present in the buffer solution.The of benzoic acid is 6.3 × 10^-5.

Accepted Answer

The pH of the solution only slightly increases upon the addition of NaOH because the added base reacts with the weak acid in the buffer solution to form its conjugate base. In this case, the weak acid is benzoic acid (C6H5COOH), and the conjugate base is sodium benzoate (C6H5COO- Na+). Therefore, the NaOH reacts with benzoic acid (D) present in the buffer solution. Since the solution is a buffer solution, the pH does not change drastically upon the addition of a small amount of base or acid.

Question 18

What is the pH of a buffer solution that is 0.172 M in hypochlorous acid (HClO)and 0.131 M in sodium hypochlorite? The  of hypochlorous acid is 3.8 ×  .

Accepted Answer

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

Question 19

Calculate the pH of a solution prepared by dissolving 0.250 mol of benzoic acid (    H)and  of sodium benzoate (Na    )in water sufficient to yield 1.00 L of solution.The  of benzoic acid is

Accepted Answer

The solution contains a weak acid (benzoic acid) and its conjugate base (sodium benzoate). We can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log([A-]/[HA])
where pKa is the acid dissociation constant of benzoic acid (4.202), [A-] is the concentration of the conjugate base (sodium benzoate), and [HA] is the concentration of the weak acid (benzoic acid).
First, we need to calculate the concentration of sodium benzoate:
moles of NaC7H5O2 = 0.250 mol
moles of NaC7H5O2 = [NaC7H5O2] x 1.00 L
[NaC7H5O2] = moles of NaC7H5O2 / 1.00 L
[NaC7H5O2] = 0.250 mol / 1.00 L
[NaC7H5O2] = 0.250 M
Since the solution was prepared by dissolving benzoic acid and sodium benzoate in water, the initial concentration of benzoic acid is also 0.250 M.
Now we can substitute the values into the Henderson-Hasselbalch equation:
pH = 4.202 + log(0.250/0.250)
pH = 4.202 + log(1)
pH = 4.202
Therefore, the pH of the solution is 4.202, which is closest to choice B (3.965).

Question 20

A buffer solution with a pH of 4.63 is prepared with  acid and  formate. The  of formic acid is

Accepted Answer

The answer of A buffer solution with a pH of...

Quiz 17: Additional Aspects of Aqueous Equilibria

What is the pH of a buffer solution that is 0.266 M in lactic acid and 0.111 M in sodium lactate? The of lactic acid is 1.4 × .

Which solution would have the greatest buffering capacity?

The of some weak acid HA is 1.76 × .The pH of a buffer prepared by combining 15.0 mL of A^- and 50.0 mL of 1.00 M HA is ________.

The of ammonia is 1.76 × .What is the pH of a buffer which is prepared by combining 50.0 mL of 1.00 M ammonia and 45.0 mL of 1.00 M ammonium nitrate?

A buffer solution with a pH of 4.31 is prepared with and The Ka of H is

0.78 M Na and are required to prepare a buffer solution with a pH of 4.40 .The of H is

What is the pH of a solution which is prepared by dissolving 0.850 mol of N and of in water sufficient to yield 1.00 L of solution? The of ammonia is

Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride.The concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is ________ M.

A buffer solution contains 0.100 M fluoride ions and 0.126 M hydrogen fluoride.What is the concentration (M) of hydrogen fluoride after addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution?

Calculate the pH of a solution prepared by dissolving 0.270 mol of weak acid HA and 0.260 mol of its conjugate base in water sufficient to yield 1.00 L of solution.The of HA is

acid and formate are required to prepare a buffer solution with a pH of 4.78 .The of formic acid is

A 25.0 mL sample of 0.150 M acetic acid is titrated with a 0.150 M NaOH solution.What is the pH at the equivalence point? The of acetic acid is 4.50 × 10^-4.

What is the pH of a solution that contains 0.800 M weak acid ( = 1.76 × ) and 0.172 M of its conjugate base?

How many milliliters of 0.0839 M NaOH are required to titrate 25.0 mL of to the equivalence point?

The addition of hydrofluoric acid and ________ to water produces a buffer solution.

What is the pH of a buffer solution that is 0.172 M in hypochlorous acid (HClO) and 0.131 M in sodium hypochlorite? The of hypochlorous acid is 3.8 × .

Calculate the pH of a solution prepared by dissolving 0.250 mol of benzoic acid ( H) and of sodium benzoate (Na ) in water sufficient to yield 1.00 L of solution.The of benzoic acid is

A buffer solution with a pH of 4.63 is prepared with acid and formate. The of formic acid is

Quiz 17: Additional Aspects of Aqueous Equilibria

What is the pH of a buffer solution that is 0.266 M in lactic acid and 0.111 M in sodium lactate? The of lactic acid is 1.4 × .

Which solution would have the greatest buffering capacity?

The of some weak acid HA is 1.76 × .The pH of a buffer prepared by combining 15.0 mL of A- and 50.0 mL of 1.00 M HA is ________.

The of ammonia is 1.76 × .What is the pH of a buffer which is prepared by combining 50.0 mL of 1.00 M ammonia and 45.0 mL of 1.00 M ammonium nitrate?

A buffer solution with a pH of 4.31 is prepared with and The Ka of H is

0.78 M Na and are required to prepare a buffer solution with a pH of 4.40 .The of H is

What is the pH of a solution which is prepared by dissolving 0.850 mol of N and of in water sufficient to yield 1.00 L of solution? The of ammonia is

Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride.The concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is ________ M.

A buffer solution contains 0.100 M fluoride ions and 0.126 M hydrogen fluoride.What is the concentration (M) of hydrogen fluoride after addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution?

Calculate the pH of a solution prepared by dissolving 0.270 mol of weak acid HA and 0.260 mol of its conjugate base in water sufficient to yield 1.00 L of solution.The of HA is

acid and formate are required to prepare a buffer solution with a pH of 4.78 .The of formic acid is

A 25.0 mL sample of 0.150 M acetic acid is titrated with a 0.150 M NaOH solution.What is the pH at the equivalence point? The of acetic acid is 4.50 × 10-4.

What is the pH of a solution that contains 0.800 M weak acid ( = 1.76 × ) and 0.172 M of its conjugate base?

How many milliliters of 0.0839 M NaOH are required to titrate 25.0 mL of to the equivalence point?

The addition of hydrofluoric acid and ________ to water produces a buffer solution.

What is the pH of a buffer solution that is 0.172 M in hypochlorous acid (HClO) and 0.131 M in sodium hypochlorite? The of hypochlorous acid is 3.8 × .

Calculate the pH of a solution prepared by dissolving 0.250 mol of benzoic acid ( H) and of sodium benzoate (Na ) in water sufficient to yield 1.00 L of solution.The of benzoic acid is

A buffer solution with a pH of 4.63 is prepared with acid and formate. The of formic acid is

The of some weak acid HA is 1.76 × .The pH of a buffer prepared by combining 15.0 mL of A^- and 50.0 mL of 1.00 M HA is ________.

A 25.0 mL sample of 0.150 M acetic acid is titrated with a 0.150 M NaOH solution.What is the pH at the equivalence point? The of acetic acid is 4.50 × 10^-4.