What should be the pH after 33 ml of 0.1 M NaOH was added to 40 ml of 0.05 M HCl?

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Answer 1 · 2017-07-03T01:39:19

12.25

$33ml(0.10M NaOH) + 40ml(0.05M HCl)$ => 1:1 Rxn Ratio

=> $0.033(0.10)"mole" NaOH + 0.040(0.05)"mole" HCl$

=> $0.0033 "mole" NaOH + 0.0020 "mole" HCl$

=> $(0.0033 - 0.0020) "mole". "of". NaOH$ in excess

=> $0.0013$ mole $NaOH$ remains in excess in 73 ml of solution.

=> $[NaOH]_(excess)$ = $[$ OH^-] $= ((0.0013"mole"OH^-)/(0.073LiterSolution))$

= $0.017 M (OH^-)$

=> $pOH = -log[OH^-] = -log(0.017) = -(-1.75) = 1.75$

=> $pH = 14 - pOH = 14 - 1.75 = 12.25$