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
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]
0.80 M
haha you already got the answer but im an idiot so good luck with that.
To find the molarity of the HCl solution, we need to use the balanced chemical equation for the reaction between HCl and KOH:
HCl + KOH -> KCl + H2O
From the equation, we can see that the mole ratio between HCl and KOH is 1:1. This means that for every 1 mole of HCl, we need 1 mole of KOH.
Given that 42.5 mL of a 0.110 M KOH solution is needed to titrate a 25.0 mL sample of the acid, we can calculate the number of moles of KOH used:
moles of KOH = concentration of KOH * volume of KOH solution
= 0.110 M * 42.5 mL
= 4.675 mmol (millimoles)
Since the mole ratio between HCl and KOH is 1:1, the number of moles of HCl in the 25.0 mL sample is also 4.675 mmol.
Now, we can calculate the molarity of the HCl solution:
molarity of HCl = moles of HCl / volume of HCl solution
= 4.675 mmol / 25.0 mL
= 0.187 M
Therefore, the molarity of the HCl solution is 0.187 M.
Hence, the correct answer is A) 0.187 M HCl.