Answer:
13.53 mL
Step-by-step 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