Question

4. Calculate the pH of a buffer solution made by adding 500 mL of 0.40 M HCl to 750 mL of 0.80 M NHs. For NH3, Kb-1.8 x 10-5. A) pH-9.95 B) pH -9.72 C)pH -9.55D) pH 9.25 E) pH-8.99

Answer 1

Given:

M(HCl) = 0.4 M

V(HCl) = 500 mL

M(NH3) = 0.8 M

V(NH3) = 750 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.4 M * 500 mL = 200 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.8 M * 750 mL = 600 mmol

We have:

mol(HCl) = 200 mmol

mol(NH3) = 600 mmol

200 mmol of both will react

excess NH3 remaining = 400 mmol

Volume of Solution = 500 + 750 = 1250 mL

[NH3] = 400 mmol/1250 mL = 0.32 M

[NH4+] = 200 mmol/1250 mL = 0.16 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.745+ log {0.16/0.32}

= 4.444

use:

PH = 14 - pOH

= 14 - 4.4437

= 9.5563

Answer: C