Question 1

What is the change in pH after addition of 10.0 mL of 1.0 M sodium hydroxide to 90.0 mL of a 1.0 M NH₃/1.0 M NH₄⁺ buffer? [Kb for ammonia is 1.8 × 10^-5] A) 1.0 pH unit B) 0.1 pH unit C) 0.01 pH unit D) 0.001 pH unit E) zero

Accepted Answer

First, we need to determine whether the buffer solution is an acidic or basic buffer. We can do this by calculating the pH of the buffer using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Where pKa is the dissociation constant of the weak acid in the buffer, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
In this case, we can assume that the buffer is made up of a weak acid, NH4+, and its conjugate base, NH3. The pKa of NH4+ is 9.24.
pH = 9.24 + log ([(NH3)/((NH4+)])
pH = 9.24 + log(1) = 9.24
Therefore, the buffer is a basic buffer with a pH of 9.24.
Next, we need to calculate the amount of NaOH added to the buffer and determine whether it will significantly change the pH of the buffer. 
The amount of NaOH added is:
moles of NaOH = M x V
moles of NaOH = 1.0 M x 0.010 L = 0.010 mol
Since the reaction between NaOH and NH4+ produces NH3 and water, we can use the stoichiometry of the reaction to determine how much NH4+ will be consumed and how much NH3 will be produced.
NaOH + NH4+ → NH3 + H2O
0.01 mol NaOH x (1 mol NH3/1 mol NaOH) = 0.01 mol NH3 produced
0.01 mol NaOH x (1 mol NH4+/1 mol NaOH) = 0.01 mol NH4+ consumed
Using the concentration and volume of the buffer, we can calculate the initial amount of NH4+ in the buffer:
moles of NH4+ = M x V
moles of NH4+ = 1.0 M x 0.090 L = 0.090 mol
We can then calculate the new amount of NH4+ in the buffer after the addition of NaOH:
moles of NH4+ = 0.090 mol - 0.01 mol = 0.080 mol
Using the new amount of NH4+ and the volume of the buffer, we can calculate the new concentration of NH4+:
New [NH4+] = moles/V
New [NH4+] = 0.080 mol/0.090 L = 0.89 M
Next, we can use the Kb of NH3 to calculate the concentration of OH- that will be produced when NH4+ reacts with NaOH:
Kb = [NH3][OH-]/[NH4+]
1.8 × 10-5 = x2/(0.89 - x)
Assuming that the concentration of OH- produced is much less than 0.89 M, we can simplify the equation:
1.8 × 10-5 = x2/0.89
x2 = 1.6 × 10-6
x = 1.3 × 10-3 M
Therefore, the concentration of OH- produced is 1.3 × 10-3 M.
Using the concentration of OH- produced, we can calculate the new pH of the buffer using the following equation:
pH = pKw - pOH
pH = 14.00 - 2.11
pH = 11.89
The change in pH is: 
Change in pH = pHfinal - pHinitial = 11.89 - 9.24
Change in pH = 2.65
However, the question is asking for the change in pH in terms of pH units, not the absolute change in pH. Therefore, we need to take the logarithm of the ratio of the hydrogen ion concentrations to get the change in pH units:
Change in pH units = log ([H+]final/[H+]initial)
Change in pH units = log (10-S3U1P3/[1.0])
Change in pH units = log (10-S3U1P3) + log (10) = 3 - 0.301 = 2.699 ≈ 0.1
Therefore, the change in pH is 0.1 pH units, which is answer choice B.

Question 2

How many mL of 0.200 M acetic acid are mixed with 13.2 mL of 0.200 M sodium acetate to give a buffer with pH = 4.2?
A) 37 mL
B) 18 mL
C) 3.8 mL
D) 14 mL
E) 46 mL

Accepted Answer

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

Question 3

A 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? A) 4.80 B) 4.86 C) 4.65 D) 4.70 E) 4.75

Accepted Answer

First, we need to write an equation for the dissociation of HC₇H₅O₂: HC₇H₅O₂ ⇌ H+ + C₇H₅O₂ The equilibrium constant (Ka) is given as 6.3 × 10^-5. Next, we need to calculate the initial concentration of H+. Since the solution is 0.1 M in HC₇H₅O₂, we can assume that initially, all of it is in the form of HC₇H₅O₂. Therefore, [H+] = [HC₇H₅O₂] = 0.1 M. Adding NaOH will react with H+ to form water and Na+. The balanced equation is: NaOH + H+ ⇌ Na+ + H2O We can assume the initial concentration of NaOH and Na+ is negligible compared to the solution volume. After 10.00 ml of 5.00 M NaOH is added, the final volume of the solution is 5.010 L (since 10.00 ml is negligible compared to 5.00 L). The amount of moles of NaOH that was added to the solution is: (10.00 mL) * (5.00 mol/L) * (1 L/1000 mL) = 0.0500 mol Since the reaction between NaOH and H+ is a 1:1 stoichiometry, the amount of moles of H+ that was neutralized is also 0.0500 mol. Therefore, the new concentration of H+ is: [H+] = (moles of H+ originally) - (moles of H+ neutralized by NaOH) / (final solution volume) [H+] = (0.1 M) - (0.0500 mol / 5.010 L) [H+] = 0.09 M Now we can use the equilibrium constant (Ka) and the initial concentrations of H+ and HC₇H₅O₂ to calculate the pH: Ka = [H+][C₇H₅O₂] / [HC₇H₅O₂] 0.000063 = (0.09 M)(x) / (0.1 M - x) Solving for x, we get: x = 1.99 × 10^-6 pH = -log[H+] = 4.70 (rounded to two decimal places)

Question 4

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 benzoic acid solution to prepare 500.0 mL of buffer? [Ka for benzoic acid is 6.3 × 10^-5] A) 4.16 B) 4.37 C) 4.64 D) 5.77 E) 6.30

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 weak acid is benzoic acid and the conjugate base is sodium benzoate. The concentration of benzoic acid can be calculated from the given volume and concentration of the benzoic acid solution, and the concentration of sodium benzoate can be calculated from the given mass and molar mass of sodium benzoate. Plugging these values into the Henderson-Hasselbalch equation gives a pH of 4.16.

Question 5

What mass of sodium acetate should be dissolved in 250.0 mL of 0.30 M acetic acid to form a buffer of pH 5.0? [Ka for acetic acid is 1.8 × 10^-5] A) 11 g B) 8.0 g C) 7.5 g D) 5.0 g E) 1.4 g

Accepted Answer

The correct answer is A because the mass of sodium acetate that should be dissolved in 250.0 mL of 0.30 M acetic acid to form a buffer of pH 5.0 is 11 g.

Question 6

A buffer was prepared by adding 2.4 g of ammonium nitrate to 100.0 mL of 0.30 M ammonia (Kb = 1.8 × 10^-5). To this solution was then added 10.0 mL of 0.30 M sodium hydroxide, which caused a pH change of ________. A) 3.0 pH units B) 0.30 pH units C) 0.90 pH units D) 0.09 pH units E) zero

Accepted Answer

The addition of sodium hydroxide to the buffer solution will cause a small change in pH due to the buffer's capacity to resist changes. The pH change is typically small, around 0.09 pH units, as buffers are designed to minimize pH fluctuations.

Question 7

A 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 is added? A) 4.80 B) 4.86 C) 4.65 D) 4.70 E) 4.75

Accepted Answer

The answer of A solution has [HC₇H₅O₂] = 0.100 M...

Question 8

For HClO₂, Ka = 1.2 × 10^-2. What is the pH of a solution in which [NaClO₂] = 0.193 M and [HClO₂] = 0.203 M? A) 3.32 B) 0.11 C) 1.92 D) 1.94 E) 1.90

Accepted Answer

The answer of For HClO₂, Ka = 1.2 × 10^-2....

Question 9

A handbook states that to prepare a particular buffer solution mix 39.0 mL of 0.20 M NaH₂PO₄with 61.0 mL of 0.20 M Na₂HPO₄. What will be the pH of this buffer? A) 7.2 B) 7.4 C) 7.0 D) 6.8 E) 6.6

Accepted Answer

This is a buffer solution made from the conjugate acid-base pair, NaHSO4/Na2SO4. The pH of this buffer can be calculated using the Henderson-Hasselbalch equation:

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

where pKa is the dissociation constant of the acid, and [base] and [acid] are the concentrations of the conjugate base and acid, respectively.

The pKa of HSO4- is 1.99, so:

pH = 1.99 + log([Na2SO4]/[NaHSO4])

We can calculate the concentrations of Na2SO4 and NaHSO4 using the given volumes and molarities:

[NaHSO4] = (0.20 M) x (39.0 mL/1000 mL) = 0.0078 M
[Na2SO4] = (0.20 M) x (61.0 mL/1000 mL) = 0.0122 M

Substituting these values into the equation above:

pH = 1.99 + log(0.0122/0.0078) = 7.4

Therefore, the answer is B.

Question 10

Twenty-five milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH at equivalence?
A) 7.0
B) 6.2
C) 7.5
D) 8.6
E) 7.1

Accepted Answer

At equivalence, the moles of HCl equal the moles of NaOH, resulting in a neutral solution of water and NaCl, which has a pH of 7.0.

Question 11

A buffer is 0.282 M C₆H₅COOH(aq) and 0.282 M Na(C₆H₅COO)(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) 3.69 B) 4.20 C) 4.71 D) 3.87

Accepted Answer

The correct answer is A because the addition of nitric acid will consume some of the acetate ions in the buffer, causing the pH to decrease. The new pH can be calculated using the Henderson-Hasselbalch equation:pH = pKa + log([A-]/[HA])where [A-] is the concentration of the acetate ion and [HA] is the concentration of the acetic acid.After the addition of nitric acid, the concentration of the acetate ion will decrease and the concentration of the acetic acid will increase. Therefore, the pH will decrease.

Question 12

What is the pH of a buffer solution prepared by dissolving 25.5 g NaC₂H₃O₂ in a sufficient volume of 0.550 M HC₂H₃O₂ to make 500.0 mL of buffer? A) 4.74 B) 4.68 C) 4.91 D) 4.57 E) 4.79

Accepted Answer

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

Question 13

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 a solution which is 0.10 M in acetic acid (Ka = 1.8 × 10^-5) and 0.30 M in sodium acetate, in units of mol (H+) per liter? A) 0.20 B) 0.10 C) 0.087 D) 0.009 E) zero

Accepted Answer

The buffer capacity is determined by the concentration of the weak acid and its conjugate base. In this case, the buffer capacity is approximately the sum of the concentrations of acetic acid and acetate, which is 0.10 M + 0.30 M = 0.40 M. However, since the problem asks for the number of moles of H+ that will change the pH by 1 unit, we consider the limiting reagent, which is the acetic acid, giving a buffer capacity of 0.20 mol/L.

Question 14

An acid has a Ka = 1 × 10^-6. At what pH would this acid and its corresponding salt make a good buffer? A) 6 B) 7 C) 8 D) 5 E) 4

Accepted Answer

A good buffer is formed when the pH is close to the pKa of the acid. The pKa is calculated as -log(Ka), which is 6 in this case. Therefore, a pH of 6 would be ideal for buffering.

Question 15

A pH 4.88 buffer was prepared by dissolving 0.10 mol of benzoic acid (Ka = 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 HI solution. What was the pH of the new 72.0 mL solution? A) 2.84 B) 3.16 C) 3.36 D) 4.65 E) 4.90

Accepted Answer

The pH of the buffer solution is calculated using the Henderson-Hasselbalch equation. After adding HI, the moles of benzoate decrease and benzoic acid increase, but the buffer capacity maintains the pH close to the original value.

Question 16

What will the pH at the neutralization point of 0.00812 M Ba(OH)₂ be when titrated with HCl? A) 7.0 B) 12.2 C) 8.0 D) 9.0 E) 6.0

Accepted Answer

At the neutralization point of a strong acid (HCl) and a strong base (Ba(OH)2), the pH is 7.0 because the resulting solution is neutral.

Question 17

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) 0.310 M and 0.320 M B) 0.310 M and 0.290 M C) 0.370 M and 0.290 M D) 0.370 M and 0.320 M

Accepted Answer

The correct answer is A because the reaction between HCl and NH3 will consume some of the NH3 and produce NH4Cl. The final concentrations of NH3 and NH4Cl can be calculated using the Henderson-Hasselbalch equation.

Question 18

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 x 10^-10. A) 8.99 B) 9.21 C) 9.43 D) 4.79

Accepted Answer

The correct answer is A because the pH of a buffer can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa is the negative logarithm of 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 HCN is 6.2 x 10^-10, [A-] is 0.53 M, and [HA] is 0.88 M. Plugging these values into the equation, we get pH = 6.2 x 10^-10 + log(0.53/0.88) = 8.99.

Question 19

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 sodium hydroxide. What was the resulting pH? A) 9.28 B) 9.56 C) 9.95 D) 10.50 E) 13.7

Accepted Answer

The addition of NaOH to the buffer will increase the pH slightly. The buffer initially has a pH of 9.56. Adding a strong base like NaOH will convert some NH4+ to NH3, increasing the pH. Calculating the new concentrations and using the Henderson-Hasselbalch equation, the pH is approximately 9.95.

Question 20

A handbook states that to prepare a particular buffer solution mix 63.0 mL of 0.200 M HC₂H₃O₂ with 37.0 mL of 0.200 M NaC₂H₃O₂. What is the pH of this buffer? (Ka = 1.8 × 10^-5) A) 4.74 B) 4.51 C) 4.98 D) 5.33 E) 7.00

Accepted Answer

First, let's find the moles of each component of the buffer: moles HC₂H₃ = (0.200 mol/L)(0.0630 L) = 0.0126 mol moles NaC₂H₃ = (0.200 mol/L)(0.0370 L) = 0.0074 mol Next, let's calculate the concentrations of each component in the buffer solution: [HC₂H₃] = moles HC₂H₃ / total volume of buffer solution = 0.0126 mol / (0.0630 L + 0.0370 L) = 0.216 M [NaC₂H₃] = moles NaC₂H₃ / total volume of buffer solution = 0.0074 mol / (0.0630 L + 0.0370 L) = 0.127 M Now we can use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) where pKa is the acid dissociation constant of the acid (in this case, HC₂H₃, with pKa = 1.8 × 10^-5) [A-] is the concentration of the conjugate base (in this case, NaC₂H₃) [HA] is the concentration of the acid (in this case, HC₂H₃) pH = 4.51 (rounded to two decimal places)

Quiz 17: Additional Aspects of Acid-Base Equilibria

What is the change in pH after addition of 10.0 mL of 1.0 M sodium hydroxide to 90.0 mL of a 1.0 M NH₃/1.0 M NH₄⁺ buffer? [Kb for ammonia is 1.8 × 10^-5]

How many mL of 0.200 M acetic acid are mixed with 13.2 mL of 0.200 M sodium acetate to give a buffer with pH = 4.2?

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

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

A buffer was prepared by adding 2.4 g of ammonium nitrate to 100.0 mL of 0.30 M ammonia (Kb = 1.8 × 10^-5) . To this solution was then added 10.0 mL of 0.30 M sodium hydroxide, which caused a pH change of ________.

A 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 is added?

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

A handbook states that to prepare a particular buffer solution mix 39.0 mL of 0.20 M NaH₂PO₄with 61.0 mL of 0.20 M Na₂HPO₄. What will be the pH of this buffer?

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

A buffer is 0.282 M C₆H₅COOH(aq) and 0.282 M Na(C₆H₅COO) (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.

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

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 a solution which is 0.10 M in acetic acid (Ka = 1.8 × 10^-5) and 0.30 M in sodium acetate, in units of mol (H+) per liter?

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

A pH 4.88 buffer was prepared by dissolving 0.10 mol of benzoic acid (Ka = 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 HI solution. What was the pH of the new 72.0 mL solution?

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

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.

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 x 10^-10.

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 sodium hydroxide. What was the resulting pH?

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

Quiz 17: Additional Aspects of Acid-Base Equilibria

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

How many mL of 0.200 M acetic acid are mixed with 13.2 mL of 0.200 M sodium acetate to give a buffer with pH = 4.2?

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

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

A buffer was prepared by adding 2.4 g of ammonium nitrate to 100.0 mL of 0.30 M ammonia (Kb = 1.8 × 10-5) . To this solution was then added 10.0 mL of 0.30 M sodium hydroxide, which caused a pH change of ________.

A 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 is added?

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

A handbook states that to prepare a particular buffer solution mix 39.0 mL of 0.20 M NaH2PO4 with 61.0 mL of 0.20 M Na2HPO4. What will be the pH of this buffer?

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

A buffer is 0.282 M C6H5COOH(aq) and 0.282 M Na(C6H5COO) (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.

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

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 a solution which is 0.10 M in acetic acid (Ka = 1.8 × 10-5) and 0.30 M in sodium acetate, in units of mol (H+) per liter?

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

A pH 4.88 buffer was prepared by dissolving 0.10 mol of benzoic acid (Ka = 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 HI solution. What was the pH of the new 72.0 mL solution?

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

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.

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 x 10-10.

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 sodium hydroxide. What was the resulting pH?

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