Question

Calculate the pH of the solution that results from each mixture.

a. 50.0mL of 0.15M HCHO2 with 75.0mL of 0.13M NaCHO2

b.125.0mL of 0.10M NH3 with 250.0mL of 0.10M NH4CL

Answer 1

a)
Concentration after mixing = mol of component / (total volume)
M(CHO2-) after mixing = M(CHO2-)*V(CHO2-)/(total volume)
M(CHO2-) after mixing = 0.13 M*75.0 mL/(75.0+50.0)mL
M(CHO2-) after mixing = 7.8*10^-2 M
Concentration after mixing = mol of component / (total volume)
M(HCHO2) after mixing = M(HCHO2)*V(HCHO2)/(total volume)
M(HCHO2) after mixing = 0.15 M*50.0 mL/(50.0+75.0)mL
M(HCHO2) after mixing = 6*10^-2 M

Ka = 1.8*10^-4

pKa = - log (Ka)
= - log(1.8*10^-4)
= 3.745

use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.745+ log {7.8*10^-2/6*10^-2}
= 3.859

Answer: 3.86

b)
Concentration after mixing = mol of component / (total volume)
M(NH3) after mixing = M(NH3)*V(NH3)/(total volume)
M(NH3) after mixing = 0.1 M*125.0 mL/(125.0+250.0)mL
M(NH3) after mixing = 3.333*10^-2 M
Concentration after mixing = mol of component / (total volume)
M(NH4+) after mixing = M(NH4+)*V(NH4+)/(total volume)
M(NH4+) after mixing = 0.1 M*250.0 mL/(250.0+125.0)mL
M(NH4+) after mixing = 6.667*10^-2 M
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 {6.667*10^-2/3.333*10^-2}
= 5.046

use:
PH = 14 - pOH
= 14 - 5.0458
= 8.9542
Answer: 8.95