How many milliliters of .45 M #HCl# will neutralize 25.0 mL of 1.00 M #KOH#?

1 Answer
anor277
Jun 14, 2016

Approx. #60*mL# of hydrochloric acid are required.

Explanation:

With all problems of this type, a BALANCED chemical equation is required to establish the equivalence:

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

Given the 1:1 equivalence:

#"Moles of KOH"=25xx10^-3*Lxx1.00*mol*L^-1=25xx10^-3*mol#

Thus we require an equivalent quantity of #0.45*mol*L^-1# hydrochloric acid:

#=# #(25xx10^-3*mol)/(0.45*mol*L^-1)xx10^3*mL*L^-1# #=# #??mL#

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