How many grams of solid KOH must be dissolved in 100 mL of a buffer solution that is 0.1 M each in weak acid HA and its potassium salt KA, to raise the pH to 6.0? (Given: pKa of HA = 5)

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Accepted Answer

0.458. pH = pKa + log([A-]/[HA]): 6 = 5 + log([A-]/[HA]) => [A-]/[HA] = 10. Initial moles (in 100 mL): HA = 0.1*0.1 = 0.01 mol, A- = 0.01 mol. Adding x mol KOH: HA + KOH -> KA + H2O. New moles: HA = 0.01-x, A- = 0.01+x. Condition: (0.01+x)/(0.01-x) = 10 => 0.01+x = 0.1-10x => 11x = 0.09 => x = 0.09/11 mol. Mass KOH = (0.09/11)*56 g = 5.04/11 = 0.458 g.

How many grams of solid KOH must be dissolved in 100 mL of a buffer solution that is 0.1 M each in weak acid HA and its potassium salt KA, to raise the pH to 6.0? (Given: pKa of HA = 5)

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