How many mL of a 3.2M NaOH solution is required to neutralize 74mL of a 12.4M HCl solution?

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Answer 1 · 2016-05-05T07:57:02

Approx. 300 mL

For all these sorts of problems we require a balanced chemical equation:

$NaOH(aq) + HCl(aq) rarr NaOH(aq) + H_2O(l)$

Such a scheme establishes that there is a 1:1 equivalence between moles acid and moles of base.

$"Moles of hydrochloric acid"$ $=$ $74xx10^-3Lxx12.4*mol*L^-1=0.895*mol$ .

Thus we require a quantity of $0.895*mol*"NaOH"$ :

$(0.895*cancel(mol))/(3.2*cancel(mol)*L^-1)$ $=$ $0.28*L" sodium hydroxide solution"$

This quantity is reasonable as the the acid is 4 times as concentrated as the base.