Question

Calculate the pH of the following solutions.

A.) 0.36 M NH4NO3

B)0.50 M Ca(NO3)2

C.)0.080 M C6H5NH2

D.) 0.45 M KNO2

Answer 1

A)

We want to get acid or base concentration so we can figure out the pH. Since NH₄NO₃ is a strong electrolyte it will break down to both components. As you should know, NH₄⁺ will release a H⁺.

K_b of NH₃ = 1.8 x 10^-5

NH₄NO₃ NH₄⁺ + NO₃^-

             NH₄⁺ NH₃ + H⁺
I            0.36........0.......0
C            -x.......... x.......x
E         0.36-x ......x.......x

K_a = K_w/K_b = (10^-14)/(1.8 x 10^-5) = 5.56 x 10^-10

K_a = [NH₃][H⁺]/[NH₄⁺] = x²/(0.36-x) = 5.56 x 10^-10

x = 1.415 10^-5 M = [H⁺]

pH = -log[H⁺] = -log(1.415 10^-5) = 4.85

C)

K_b of C₆H₅NH₂ = 3.8 x 10^-10

          C₆H₅NH₂ C₆H₅NH^- + H⁺
I          0.080.................0.........0
C            -x....................x.........x
E       0.080-x ...............x.........x

K_a = K_w/K_b = (10^-14)/(3.8 x 10^-10) = 2.63 x 10^-5

K_a = [NH₃][H⁺]/[NH₄⁺] = x²/(0.080-x) = 2.63 x 10^-5

x = 1.45 10^-3 M = [H⁺]

pH = -log[H⁺] = -log(1.45 10^-3) = 2.84

D)

KNO₂ is a strong salt : KNO₂ K⁺ + NO₂^-

NO₂^- + H₂O HNO₂ + OH^-

K_a of HNO₂ = 4.3 10^-4
K_b = K_w/K_a = (10^-14)/(4.3 x 10^-4) = 2.32 10^-11

2.32 10^-11 = x²/ 0.45-x
x = [OH-]= 3.23 10^-6 M
pOH = 5.49
pH = 14 - 5.49 = 8.51