What volume of 0.250 M nitric acid is needed to neutralize 17.35 mL of 0.195 M KOH solution?

13.53 mL

Explanation:

Step 1:

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

HNO3 + KOH —> KNO3 + H2O

From the equation above, we obtained the following information:

Mole ratio of the acid (nA) = 1

Mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question. This includes:

Volume of acid (Va) =?

Molarity of the acid (Ma) = 0.250 M

Volume of base (Vb) = 17.35 mL

Molarity of the base (Mb) = 0.195 M

Step 3:

Determination of the volume of the nitric acid needed for the reaction. The volume of nitric acid needed for the reaction can be obtained as follow:

MaVa/MbVb = nA/nB

0.250 x Va/ 0.195 x 17.35 = 1

Cross multiply

0.250 x Va = 0.195 x 17.35

Divide both side by 0.250

Va = (0.195 x 17.35) /0.250

Va = 13.53 mL

Therefore, the volume of nitric acid needed for the reaction is 13.53 mL

i'm assuming that the correct answer is indeed the first one/ "a"..

a corroding iron, if that's one of your choices