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Calculate the pH of a solution which contains 9.9 mlof 1 M HCI and 100 ml of 0.1 M MaoH Calculate the pH of a solution which contains 9.9 mlof 1 M HCI and 100 ml of 0.1 M MaoH
Dear Student Na OH reacts with H Cl : Na OH + H Cl → Na Cl + H2O 1 mol Na OH reacts with 1 mol HCl Mol H Cl in 9 . 9 m L of 1 . 0 M solution = 9.9 /1000 * 1.0 = 0.0099 mol H Cl Mol Na OH in 100mL of 0.1 M solution = 100 / 1000 * 0.1 = 0.01 mol Na OH On mixing the 0.0099 mol H Cl will be neutralized and 0.01 - 0.0099 = 1 * 10 ^ - 4 mol Na OH dissolved in 109.9 m L solution Morality of Na OH solution = ( 1 * 10 ^ - 4 ) / 0.1099 = 9.099 * 10 ^ - 4 M p OH = - log ( 9.099 * 10 ^ - 4) p OH = 3.04 pH = 14.00 – p OH pH = 14.00 - 3.04 pH = 10.96 RegardsArun (AskIITians Forum Expert)
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