1. calculate the ph of a 2.5m solution of nh3 (kb = 1.8 x 10-5 ).

11.83.

Explanation:

∵ pOH = - log[OH⁻]

∵ [OH⁻] = √(Kb.C),

Kb = 1.8 x 10⁻⁵, C = 2.5 M.

∴ [OH⁻] = √(Kb.C) = √(1.8 x 10⁻⁵)(2.5 M) = 6.71 x 10⁻³ M.

∴ pOH = - log[OH⁻] = - log(6.71 x 10⁻³ M) = 2.17.

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 2.17 = 11.83.

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