Question

Using activities, calculate the pH of a solution containing 0.010 M NaOH plus 0.0120 M LiNO3.

Answer 1

Sol:-

The activities (μ) of given salts will be calculated as :

μ = 0.5(0.010 M Na⁺ + 0.010 M OH^- + 0.0120 M Li⁺ + 0.0120 M NO₃^-)
μ = 0.022 M
Kw = γH⁺[H⁺]·γOH^-[OH^-]

Thus,

γH⁺[H⁺]= Kw/(γOH^-[OH^-]) = 1.0 x 10^-14/(γOH^- x 0.010 M)

According to Debye-Huckel equation:
logγOH^- = [ (-0.51)(-1)²·(0.022)^0.5] / [1+(350) (0.022)^0.5/305] = - 0.06587

γOH^- = 10^{- 0.06587}
γOH^- = 0.8593
γH⁺[H⁺] = 1.0 x 10^-14/ (0.8593 x 0.010) = 1.16 x 10^-12 M

We know pH = -log(γH⁺[H⁺])

pH = - log (1.16 x 10^-12 )

pH = - (- 11.94 )

pH = 11.94