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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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

Explanation:

$33 m l (0.10 M N a O H) + 40 m l (0.05 M H C l)$ $33ml(0.10M NaOH) + 40ml(0.05M HCl)$ => 1:1 Rxn Ratio

=> $0.033 (0.10) mole N a O H + 0.040 (0.05) mole H C l$ $0.033(0.10)"mole" NaOH + 0.040(0.05)"mole" HCl$

=> $0.0033 mole N a O H + 0.0020 mole H C l$ $0.0033 "mole" NaOH + 0.0020 "mole" HCl$

=> $(0.0033 - 0.0020) mole . of . N a O H$ $(0.0033 - 0.0020) "mole". "of". NaOH$ in excess

=> $0.0013$ $0.0013$ mole $N a O H$ $NaOH$ remains in excess in 73 ml of solution.

=> ${[N a O H]}_{e x c e s s}$ $[NaOH]_(excess)$ = $[$ $[$ OH^-] $= (\frac{0.0013 mole O H^{-}}{0.073 L i t e r S o l u t i o n})$ $= ((0.0013"mole"OH^-)/(0.073LiterSolution))$

= $0.017 M (O H^{-})$ $0.017 M (OH^-)$

=> $p O H = - log [O H^{-}] = - log (0.017) = - (- 1.75) = 1.75$ $pOH = -log[OH^-] = -log(0.017) = -(-1.75) = 1.75$

=> $p H = 14 - p O H = 14 - 1.75 = 12.25$ $pH = 14 - pOH = 14 - 1.75 = 12.25$

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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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