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Use the Henderson eqn pOH = pKb + log(baseacid).pdf

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Use the Henderson eqn: pOH = pKb + log(base/acid) pOH = -log(1.88 x 10^-5) + log(.750/.250) pOH = 5.20 pH + pOH = 14 pH + 5.20 = 14 pH = 8.8 Solution Use the Henderson eqn: pOH = pKb + log(base/acid) pOH = -log(1.88 x 10^-5) + log(.750/.250) pOH = 5.20 pH + pOH = 14 pH + 5.20 = 14 pH = 8.8.

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Use the Henderson eqn: pOH = pKb + log(base/acid) pOH = -log(1.88 x 10^-5) + log(.750/.250) pOH = 5.20 pH + pOH = 14 pH + 5.20 = 14 pH = 8.8 Solution Use the Henderson eqn: pOH = pKb + log(base/acid) pOH = -log(1.88 x 10^-5) + log(.750/.250) pOH = 5.20 pH + pOH = 14 pH + 5.20 = 14 pH = 8.8

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  • 1. Use the Henderson eqn: pOH = pKb + log(base/acid) pOH = -log(1.88 x 10^-5) + log(.750/.250) pOH = 5.20 pH + pOH = 14 pH + 5.20 = 14 pH = 8.8 Solution Use the Henderson eqn: pOH = pKb + log(base/acid) pOH = -log(1.88 x 10^-5) + log(.750/.250) pOH = 5.20 pH + pOH = 14 pH + 5.20 = 14 pH = 8.8