Question

 

1) What is the pH of a buffer solution that is 0.255 M in hypochlorous acid (HClO) and 0.333 M in sodium hypochlorite? The Ka of hypochlorous acid is 3.8 × 10-8.

2) What is the pH of a solution prepared by mixing 25.00 mL of 0.10 M CH3CO2H with 25.00 mL of 0.040 M CH3CO2Na? Assume that the volume of the solutions are additive and that Ka = 1.8 × 10-5 for CH3CO2H.

3) What is the pH of a buffer system prepared by dissolving 10.70 grams of NH4Cl and 35.00 mL of 12 M NH3 in enough water to make 1.000 L of solution? Kb = 1.80 × 10-5 for NH3.

Answer 1

1)

For a buffer solution that consists of a weak acid and its conjugate base, buffer pH can be calculated by using Henderson-Hasselbalch equation.

In this problem, the buffer consists of HClO, a weak acid and its conjugate base, ClO^-.

Sodium hypochlorite dissociates in 1:1 ratio to give hypochlorite anion, ClO^-.

So, [NaClO] = [ClO^-] = 0.333 M

Calculate the pKa of the hypochlorous acid.

pKa = -log Ka = - log(3.8*10^-8) = 7.42

Now, substitute the given values in the above given Henderson-Hasselbalch equation.

Hence, the pH of the buffer solution is 7.30

2)

Calculate the moles of CH₃COOH and CH₃COONa.

0.025 L * 0.10 mol/L = 0.0025 mol CH₃COOH

0.025 L * 0.040 mol/L = 0.001 mol CH₃COONa

Total volume = 50 mL = 0.050 L

Now, calculate the molarities.

[CH₃COOH] = 0.0025 mol / 0.050 L = 0.05 M

[CH₃COONa] = [CH₃COO^-] = 0.001 mol / 0.050 L = 0.02 M

Calculate pKa

pKa = -log Ka = - log(1.8*10^-5) = 4.74

Now, calculate the pH using the Henderson-Hasselbalch equation.

Therefore, the pH of the solution is 4.34

3)

First we need to calculate the moles of each species. Ammonia is the base and ammonium ion is the acid.

NH₃ = 0.035 L * 12 mol/L = 0.42 mol

NH₄Cl = 10.70 g * 1mol / 53.49 g = 0.20 mol

Now, calculate the molarities. The volume of buffer is 1 L, so the molarities of both the solutions would be the same as their number of moles.

Hence, [NH₃] = 0.42 M

[NH₄⁺] = 0.20 M

Ka and kb are related by following equation:

pKa = 14 - pKb

pKb = -log(1.80*10^-5) = 4.74

Thus, pKa = 14 - 4.74 = 9.26

Now, substitute the values in Henderson-Hasselbalch equation to calculate the pH.

Therefore, the pH of the buffer system is 9.58