Suppose that 10.1 mL of HNO3 is neutralized

by 61.9 mL of a 0.0041 M solution of KOH

in a titration. Calculate the concentration of

the HNO3 solution.

Answer in units of M.

0.0251 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

HNO3 + KOH - > KNO3 + H2O

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Next, the data obtained from the question. This include the following:

Volume of acid (Va) = 10.1mL

Molarity of acid (Ma) = ... ?

Volume of base (Vb) = 61.9 mL

Molarity of base (Mb) = 0.0041 M

Next, we shall determine the molarity of the acid, HNO3 as follow:

MaVa/MbVb = nA/nB

Ma x 10.1 / 0.0041 x 61.9 = 1

Cross multiply to express in linear form

Ma x 10.1 = 0.0041 x 61.9

Divide both side by 10.1

Ma = (0.0041 x 61.9) / 10.1

Ma = 0.0251 M

Therefore, the molarity of HNO3 is 0.0251 M