# 1.553g of a monoprotic acid (MM = 97.10 g/mole) dissolved in 75.00mL of D.I. water was titrated with KOH. At the endpoint, 35.28 mL of KOH was consumed. Calculate the molarity of KOH?

Feb 18, 2015

The molarity of the $K O H$ solution will be $\text{0.454 M}$.

When performing a titration, you determine the end-point by observing the color change of the solution. In theory, the end-point and the equivalence point should be the same, but you can get discrepancies in actual experiments.

What the end-point tells you is that the acid and the base are now mixed in "exact" proportions to neutralize each other, i.e. to cancel each other out.

When that happens, the number of moles of acid is equal to the number of moles of $K O H$. The number of moles of acid you have in solution is

$\text{1.553 g" * ("1 mole")/("97.10 g") = "0.0160 moles}$

SInce the number of moles of $K O H$ must be equal to this value, you'll get

$\text{0.0160 moles acid" * ("1 mole KOH")/("1 mole acid") = "0.0160 moles KOH}$

The molarity of the added $K O H$ solution will be

${C}_{K O H} = \frac{n}{V} = \text{0.0160 moles"/(35.28 * 10^(-3)"L") = "0.454 M}$