What volume of a 1.0 M HCI is required to completely neutralize 25.0 ml of a 1.0 M KOH?

1 Answer
anor277
Apr 7, 2017

Why, an equal volume, i.e. a #25.0*mL# volume............

Explanation:

We interrogate the reaction:

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

When we quote solution concentration, we refer to the quotient:

#"Concentration"# #=# #"Moles of solute"/"Volume of solution"#, i.e.

#C=n/V#, and thus #"concentration"# has units of #mol*L^-1#

And you will habitually use this quotient to calculate to calculate solution concentrations, moles and mass of solute, and volume of solution needed.

It should not take a great deal of insight to appreciate that the given volume of #NaOH(aq)# solution contains THE SAME AMOUNT OF SOLUTE that is in the given volume of #HCl(aq)#.

i.e. #"moles of KCl"# #=#

#"concentration"xx"volume"=1.0*mol*L^-1xx25.0xx10^-3*L#

#=25xx10^-3*mol#.

What is the final concentration of #KCl(aq)#? Why is it different from the starting concentrations of #[KOH]#, and #[HCl]#?

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