Question

If the Kb of a weak base is 8.7 × 10-6, what is the pH of a 0.33 M solution of this base?

Answer #1

Construct the ICE table

water is in the large quantity need not consider that term

B + H2O <--------> BH + OH^{-}

I 0.33 0 0

C -x +x +x

E 0.33-x +x +x

write the Kb expression

Kb = [OH-][BH] / [B]

8.7 × 10^{-6} = [x][x] / 0.33-x]

x^{2} + 8.7 ×
10^{-6}x -2.871 x 10^{-6}

solve the quadratic equation

x = 0.00169 M = [OH-] = [BH]

now we know the concentration of [OH-]

pOH = -log[OH-]

pOH = -log[0.00169]

pOH = 2.77

now use the formula

pH + pOH = 14

pH = 14-pOH

pH = 14-2.77

pH = 11.23

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