Question

Please answer all numerically with three demical places. :)

A) What is the pH of a buffer prepared by adding 0.405 mol of the weak acid HA to 0.609 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7.

B) What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid.

D) What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.

Answer 1

A) What is the pH of a buffer prepared by adding 0.405 mol of the weak acid HA to 0.609 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7.

Solution :- Using the Henderson equation we can calculate the pH of the buffer solution

Initial molarity of the acid = 0.405 mol / 2.0 L = 0.2025 M

Initial molarity of the base = 0.609 mol / 2 L = 0.3045 M

pH= pka + log [base /acid]

pka= - log ka

pka = -log 5.66*10^-7

pka = 6.247

now lets calculate the pH

pH= pka + log [base /acid]

pH= 6.247 + log [0.3045 /0.2025]

pH= 6.424

B) What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid.
Solution :- When the 0.150 mol HCl is added then it reacts with the base and produces acid

Therefore the moles of base remain after reaction = 0.609 mol – 0.150 mol = 0.459 mol

Moles of acid present after reaction = 0.405 mol +0.150 mol = 0.555 mol

New molarities of the acid and base are

[base] = 0.459 mol / 2.0 L = 0.2295 M

[acid] = 0.555 mol/2.0 L = 0.2775 M

Now lets calculate the pH

pH = pka + log [base/ acid]

pH=6.247 + log [0.2295/0.2775]

pH= 6.165

  D) What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.

Solution :- When base is added to the buffer then it reacts with acid and forms the conjugate base

So the moles of acid remain after reaction = 0.405 mol – 0.195 mol = 0.210 mol

Moles of base after the reaction = 0.609 mol + 0.195 mol = 0.804 mol

New molarities are as follows

[acid ]= 0.210 mol /2.0 L = 0.105 M

[base] = 0.804 mol / 2.0 L = 0.402 M

Now lets calculate the pH

pH= pka + log [base / acid]

pH= 6.247 + log [0.402/0.105]

pH= 6.830