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Calculate the pH of a solution which contains 9.9 ml of 1 M HCI and 100 ml of 0.1 M NaOH |
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Answer» NaOH reacts with HCl : NaOH + HCl → NaCl + H2O 1 mol NaOH reacts with 1 mol HCl Mol HCl in 9.9mL of 1.0M solution = 9.9/1000*1.0 = 0.0099 mol HCl Mol NaOH in 100mL of 0.1M solution = 100/1000*0.1 = 0.01 mol NaOH On mixing the 0.0099 mol HCl will be neutralised and 0.01 - 0.0099 = 1*10^-4 mol NaOH dissolved in 109.9mL solution Molarity of NaOH solution = (1*10^-4) / 0.1099 = 9.099*10^-4M pOH = -log ( 9.099*10^-4) pOH = 3.04 pH = 14.00 - pOH pH = 14.00 - 3.04 pH = 10.96 |
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