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Calculate the [OH-] of a solution whose pH is 9.62. |
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Answer» pH + pOH = 14 pOH = 14 – pH = 14 – 9.62 = 4.38 pOH = 4.38 = – log10 [OH– ]; 4.38 = – log10 [OH– ] log10[ OH– ] = – 4.38 ; [OH–] = antilog (–4.38) = antilog (5 – 4.38 – 5) = antilog (0.62 – 5) = antilog (0.62) x 10–5 = 4.169 × 10–5 mol dm–3. |
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