Question

Determine the pH of each of the following two-component solutions.

0.135 M in NH4Cl and 0.150 M in HNO3

8.92×10⁻² M in sodium benzoate and 0.220 M in potassium bromide

Answer 1

NH4Cl --> NH4+ and Cl-

NH4+ +H2O <--> NH3 + H3O+

note Ka is pretty small

then assume only HNO3 interaction

p H= -log(H) = -log(0.15) =0.8239

b)

NaBenzoate + KBr

KBr has no effect

but NaBenzoate

Benzoate + H2O <--> HBenzoic + OH-

Kb = [HBenzoic ]OH-]/[Benzoate ]

Ka = 6.5*10^-5

Kb = (10^-14)/(6.5*10^-5) = 1.538*10^-10

Kb = [HBenzoic ]OH-]/[Benzoate ]

1.538*10^-10 = x*x/(8.92*10^-2 - x)

solve for x

x = [OH-] = 0.0000037037049518257268

pOH = .og(OH) = -log(0.0000037037049518257268) =5.431

pH = 14-pOH =14-5.431 = 8.569