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What is pH of mixture prepared by mixing 100 mL 0.1 M NaOH and 9.9 mL 1 M HCl?
100ml .1M NaOH and 9.9 ml 1M HClNow nonof moles of NaoH in solutions will be 100(.1) = 10 mili moles that gives 10 mol of OH- ions in solution. Now 9.9 ml of 1 M i.e. total of 9.9 mmols of H+ ions Now 9.9 mmol will react will 10 mmol to give .1 mmol of OH- ions in total of (100+9.9=109.9) ml solution. So the concentration of OH- ions will be (.1mmol/109.9) =.0009099 or .00091 mmolar And H+ X OH- =10^(-14)H+=10-14/{9.1*10^-4*10^-3)= 10^-7/9.1=1.098*10^-8 so the pH=-log( H+) =7.96
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