What is the molarity of an HCl solution if 25.0 mL of this solution required 17.80 mL of 0.108 M NaOH to reach the end point in a titration?

Explanation:

We are asked to find the molarity of an acid given the details of a titration experiment. The formula for titration is as follows:

In this formula, M is the molarity of the acid or base and V is the volume of the acid or base. The molarity of the hydrochloric acid (HCl) is unknown and the volume is 25.0 milliliters.

The molarity of the sodium hydroxide (NaOH) is 0.108 molar and the volume is 17.80 milliliters.

We are solving for the molarity of the acid and we must isolate the variable . It is being multiplied by 25.0 milliliters. The inverse operation of multiplication is division, so we divide both sides of the equation by 25.0 mL.

The units of milliliters cancel.

The original measurements have 3 and 4 significant figures. We must round our answer to the least number of sig figs, which is 3. For the number we calculated, that is the ten-thousandth place. The 9 to the right of this place tells us to round the 8 up to a 9.

The molarity of the hydrochloric acid is 0.0769 Molar.