Question

A hypothetical weak acid, HA, was combined with NaOH in the following proportions: 0.20 mol of HA, 0.080 mol of NaOH. The mixture was diluted to a total volume of 1.0 L and the pH measured.

(a) If pH = 4.80, what is the pKa of the acid?

(b) How many additional moles of NaOH should be added to the solution to increase the pH to 5.00?

Answer 1

(a)

Weak acid HA combined with NaOH:

 

HA(aq) → H+(aq) + A-(aq)

Initially 0.20 mol HA and 0.080 mol of NaOH present Thus after the reaction 0.12 mol of HA and 0.080 mol present in 1L solution

 

[HA] = 0.12M and [A-] = 0.080M

 

To calculate [H+]:-:

[H+] = antilog of - pH

         = anti log of -4.80

         = 1.585 × 10-5 M

 

To calculate Ka:

Ka = [H+][A-]/[HA]

Ka = (1.585 × 10-5)(0.080)/(0.12)

Ka = 1.057 × 10-5

 

To calculate pKa:

pKa = -log Ka

         = -log 1.057 × 10-5

         = 4.976

pKa ≈ 4.98

 

(b) To calculate concentration of H+ ion when pH = 5.00

[H+] = antilog of - pH

        = antilog of – 5.00

[H+] = 1.0 × 10-5 M

 

To calculate extra moles of NaOH:

 

Let x be the extra moles of NaOH

 

Ka = [H+][A-]/[HA]

⇒ 1.057 × 10-5 = (1.0 × 10-5)(0.0800 + x)/(0.12 – x)

⇒ 1.057 × 10-5(0.12 – x) = (1.0 × 10-5)(0.080 + x)

⇒ 1.268 × 10-5(0.12 – x) = (1.0 × 10-5)(0.080 + x)

⇒ 4.680 × 10-7 = 2.057 × 10-5x

⇒ x = 4.680 × 10-7/2.057 × 10-5

⇒ x = 2.275 × 10-2

⇒ x = 0.02275 mol

⇒ x ≈ 0.023 mol

 

0.023 mol should be added to increase pH to 5.00.