How many mL of 0.20 M hydrochloric acid is required to neutralize 100 mL of 0.80 M potassium hydroxide?

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Answer 1 · 2016-08-11T15:44:30

$"Volume of HCl"=400$ $mL$ .

$HCl(aq) + KOH(aq) rarr KCl(aq) + H_2O(l)$

Now the reaction is trivial here. It is nevertheless good practice to write the stoichiometric equation to show the $1:1$ equivalence.

$"Moles of KOH"$ $=$ $0.80*mol*L^-1xx0.100*L$ $=$ $0.080*mol$ . Note that here I normalized the volumes, $1*mL$ $=$ $10^-3L$ .

Thus $"volume of HCl"$ $=$ $(0.080*mol)/(0.20*mol*L^-1)$ $=$ $0.40*L$

PS Sorry for shouting in the top answer; I still haven't got the hang of this editor.