Question

determine the pH of 263 ml of solution which has NH4I = 0.300 M Kb = 1.74x10^-5 for NH3(aq). show work.

Answer 1

Sol :-

NH₄HI is the salt of weak base (i.e. NH₄OH) and strong acid (i.e. HI) . On hydrolysis NH₄I gives a weak base and strong acid as :

NH₄I (aq) + H₂O (l) <-------> NH₄OH (aq) + HI (aq)

We know that the pH of salt of weak base and strong acid can be calculated by using equation i.e.

pH = 1/2 [ pK_w - pK_b - logC ] ..............(1)

here, K_w = ionic product of water = 10^-14 at 25⁰C

we know

pK_w = - log K_w

so

pK_w = - log 10^-14

pK_w = - ( -14 )

pK_w = 14

given dissociation constant for base i.e. K_b = 1.74 x 10^-5

we know

pK_b = - log K_b

pK_b = - log 1.74 x 10^-5

pK_b = - ( - 4.76 )

pK_b = 4.76

also

C = Concentration of salt = 0.300 M (given)

Now substitute all these values in equation (1), we have

pH = 1/2 [ 14 - 4.76 - log 0.300 ]

pH = 1/2 [ 14 - 4.76 - (- 0.523) ]

pH = 1/2 [ 14 - 4.76 + 0.523 ]

pH = 1/2 [ 9.763]

pH = 4.88

Hence the pH = 4.88