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

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation:

your answer would be “dry”