Answer:
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
