What is the pH of a solution formed by mixing 125.0 mL of a .0250 M HCL and 75.0

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Explanation / Answer

HCl reacts with KOH to form salt and water. HCl + KOH => KCl + H2O KCl does not undergo salt hydrolysis. Hence pH of KCl is not taken into consideration. As you can see, the ratio of number of moles of HCl to KOH reacted is 1:1. Hence we can see that KOH is in excess. First, find the number of moles of KOH in excess => (50/1000 x 0.125) - (25/1000 x 0.125) = 0.003125 mol Since KOH is a strong base, [OH-] = [KOH] (Square brackets denote concentration) Therefore, [OH-] = 0.003125 / ((25+50)/1000) = 0.0417 (3 sf) pOH= -lg[OH-] = 1.38 (3sf) We also know that pOH + pH=14. Hence, pH=14-1.38=12.6 (1dp) Hope this helps you (:

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