Question

Calculate the pH of the solution that results from each of the following mixtures.

140.0 mL of 0.27 M HF with 220.0 mL of 0.32 M NaF

165.0 mL of 0.11 M C2H5NH2 with 270.0 mL of 0.22 M C2H5NH3Cl

Answer 1

1)

Ka = 6.8 x 10^-4

Moles of HF = 140 x 0.27 / 1000 = 0.0378

Moles of NaF = 220 x 0.32 / 1000 = 0.0704

Total Volume = 140 + 220 = 360 ml

Concentration of F^- = [F^-] = ( 0.0378 + 0.0704 ) / ( 360 / 1000 ) = 0.3005 M

2)

K_b = 1.3 x 10^-3

Moles of C₂H₅NH₂ = 165 x 0.11 / 1000 = 0.01815

Moles of C₂H₅NH₃⁺ = 270 x 0.22 / 1000 = 0.0594

Total Volume = 165 + 270 = 435 ml

Concentration of C₂H₅NH₃⁺= [C₂H₅NH₃⁺] = ( 0.01815 + 0.0594 ) / ( 435 / 1000 ) = 0.1783 M