Question

Calculate all equilibrium concentrations and the pH of each of the following solutions. a. 0.20 M NaCHO_2 (sodium formate) b. 0.20 M NaC_7 H_5 O_2 (sodium benzoate) c. 0.20 M NaNO_3 (Partial answers: (a) pH = 8.52: (b) pH = 8.74: (c) pH = 7.00)

Answer 1

a

Kb of freaction is =  5.9x10^-11

let conc [OH] = x

5.9x10^-11 = x^2/(.2-x)

this gives x= 3.43514×10^-6and other value is negative so it is not considered

pOH= -log(3.43514×10^-6) = 5.48

pH= 14 - 5.48= 8.52

part b

K for the reaction is 1.547*10^-10

let conc [OH] = x

1.547*10^-10 = x^2/(.2-x)

we get x= 5.5623×10^-6

pOH = -log(5.5623×10^-6) = 5.26

pH = 14 -5.26 = 8.74

part c

NaNO3 is a salt of strong acid (HNO3) and strong base (NaOH). always such a reaction produces nuetral salt.

So pH of NaNO3 sol will always be 7 irrespective of its concentration.