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Calculate the [OH] of a solution whose pH is 9.62. |
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) × 10-5 = 4.169 × 10-5 mol dm-3 |
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