Part A) What is the pH of a buffer prepared by adding 0.809 mol of the weak acid

Question

Part A)

What is the pH of a buffer prepared by adding 0.809 mol of the weak acid HA to 0.305 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×107. Express the pH numerically to three decimal places.

Part 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.

Express the pH numerically to three decimal places.

Part C)

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.

Express the pH numerically to three decimal places.

Explanation / Answer

A. Ka = 5.66*10-7

pKa = -logKa = -log5.66*10-7 = 6.25

pH = pKa + log[NaA]/[HA]

pH = 6.25 + log0.305/0.809

pH = 6.25-0.42

pH = 5.83

Part b- When 0.150mol HCl is added-

HCl being a strong acid will completely react with the conjugate base so-

HCl + NaA--> HA + NaCl

So moles of HA = 0.809mol + 0.150mol = 0.959moles

Moles of NaA = 0.305mol - 0.150mol = 0.155moles

pH = 6.25 + log0.155/0.959

pH = 6.25 - 0.79 = 5.46

Part C -

When 0.195 mol of NaOH is added it reacts with the weak acid present in the buffer and gives the conjugate base

NaOH + HA ---> NaA + H2O

So moles of HA = 0.809mol - 0.195mol = 0.614mol

Moles of NaA = 0.305mol + 0.195mol = 0.5mol

pH = 6.25 + log0.5/0.614

pH = 6.25 - 0.089 = 6.16

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