Question

PART A: What is the pH of a buffer prepared by adding 0.506 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 5.66×10−7?

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.

Please explain how to get the answer.

Answer 1

PART A.

A buffer solution resists changes in pH when it is diluted or when acids or bases are
added to it.

The Henderson-Hasselbalch approximation allows us one method to approximate the pH of a buffer solution.

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

Given Ka = 5.66×10⁻⁷

We have to find the concentration

Concentration = Number of moles / Volume

[NaA] = 0.305 mol / 2.00 L   = 0.1525 M

[HA] = 0.506 / 2.00 L = 0.253 M

pKa = -log Ka = -log(5.66×10⁻⁷) = 6.247

pH = 6.247 + log( 0.1525 / 0.253 )

pH = 6.027

So, the pH of Buffer is 6.027

PART B.

So when HCl is added to the bufffer, it reacts with base component of the buffer A- as

                  HCl    + A- <------->      HA

As a result the concentration of HA increases and A- decreases.

Number of moles of HCl added is 0.150.

Number of moles of NaA left after reacting with HCl is   = 0.305 - 0.150 = 0.155 Mol

New concentration after reaction is    0.155 mol / 2.00 L (Since volume change is negligible)

= 0.0775 M

Number of moles of HA left after reacting with HCl is   = 0.506 + 0.150 = 0.656 Mol

New concentration after reaction is    0.656 mol / 2.00 L (Since volume change is negligible)

= 0.328 M

Substituting the new concentration in Henderson-Hasselbalch equation we have

pH = 6.247 + log( 0.0775 / 0.328 )

pH = 5.620

PART C.

Likewise when NaOH is added to the bufffer, it reacts with acid component of the buffer H+ as

                  NaOH    + HA <------->      H₂O + NaH

As a result the concentration of HA decreases and A- Increases.

Number of moles of NaOH added is 0.195.

Number of moles of NaA left after reaction is   = 0.305 + 0.195 = 0.5 Mol

New concentration after reaction is    0.5 mol / 2.00 L (Since volume change is negligible)

= 0.25 M

Number of moles of HA left after reaction is   = 0.506 - 0.195 = 0.311 Mol

New concentration after reaction is    0.311 mol / 2.00 L (Since volume change is negligible)

= 0.155 M

Substituting the new concentration in Henderson-Hasselbalch equation we have

pH = 6.247 + log( 0.25 / 0.155 )

pH = 6.454