Question

10). What is the molarity of an HCl solution if 42.5 mL of a 0.110 M KOH solution is needed to titrate a 25.0-mL sample of the acid?
A) 0.187 M HCl
B) 0.587 M HCl
C) 0.0647 M HCl
D) 5.35 M HCl
E) 1.70 M HCl

Answer 1

from the equation

HCl + KOH -> KCl + H2O

1 mole of HCl reacts with 1 mole of KOH

Then if we use

M1V1=M2V2

42.5 ml x 0.110 M = 25.0 ml x X

X=42.5 ml x 0.110 ml/25.0 ml

[Incidently you can use 'school boy logic' here to eliminate amswers. If we only have 25.0 ml and we are titrating against 42.5 ml of 0.110 M then the 25.0 ml solution must be stronger (higher molarity) than 0.110 M. so it can't be (C).
As the ratio of volumes is 42.5 to 25.0 then we expect the answer to be about 1.5 times more concentrated than 0.110 M, not
(B) which is 5x
(D) is 50x and
(E) is 17x,
so the answer is (A) without touching a calculator]

Answer 2

0.80 M

Answer 3

haha you already got the answer but im an idiot so good luck with that.