Question

What is the pH of the solution which results from mixing 50.0 mL of 0.30 M HF(aq) and 50.0 mL of 0.30 M NaOH(aq) at 25 degrees celcius? (Ka of HF = 7.2*(10^-4)

Answer 1

HF:

Molarity = 0.30 M

Volume = 50.0 mL = 0.05 L

Moles of HF = Molarity x volume =  0.30 M x 0.05 L = 0.015 moles

NaOH:

Molarity = 0.30 M

Volume = 50.0 mL = 0.05 L

Moles of NaOH = Molarity x volume =  0.30 M x 0.05 L = 0.015 moles

Since HF is weak acid and NaOH is a strong base,

HF + OH ^- --------------> F^- + H2O

moles of F^- = 0.015 moles

total volume = 50.0 mL + 50.0 mL = 100 mL = 0.1 L

[F-] = moles / volume= 0.015 moles/ 0.10 L = 0.15 M

F^- + H2O ---------------> HF + OH^-

0.15 M 0 0

0.15 -x x x

Kb = [F] [OH-] / [F^-]

Kw/Ka = x.x / (0.15 -x) ( Kw = Ka.Kb)

Since x is very small, 0.15-x = x,

Therefore, Kw/Ka = x.x / (0.15)

1.0 x 10^-14/7.2x10^-4 =x²/ (0.15)

0.139 x 10^-10 = x² / (0.15 )

  x² = (0.139 x 10^-10)(0.15 )

x = 0.1444 x 10^-5

Therefore,

[OH-] = 0.1444 x 10^-5 M

Hence,

pOH = -log [OH-]

= - log (0.1444 x 10^-5)

= 5.84

pOH = 5.84

pH = 14 - pOH = 14 - 5.84 = 8.16

Therefore, pH = 8.16