Question

When 20.00 mL of an unknown monoprotic acid is titrated with 0.125 M NaOH, it takes 15.00 mL to reach the endpoint. What is the molarity of the unknown acid?

Answer 1

About 0.0940 M.

Step-by-step explanation:

Recall that NaOH is a strong base, so it dissociates completely into Na⁺ and OH⁻ ions. Because the acid is monoprotic, we can represent it with HA. Thus, the reaction between HA and NaOH is:

`\displaystyle \text{HA}_\text{(aq)} + \text{OH}^-_\text{(aq)} \longrightarrow \text{H$_2$O}_\text{($\ell$)} + \text{A}^-_\text{(aq)}`

Using the fact that it took 15.00 mL of NaOH to reach the endpoint, determine the number of HA that was reacted with:


`\displaystyle \begin{aligned} 15.00\text{ mL} &\cdot \frac{0.125\text{ mol NaOH}}{1\text{ L}} \cdot \frac{1\text{ L}}{1000\text{ mL}} \\ \\ &\cdot \frac{1\text{ mol OH}^-}{1\text{ mol NaOH}} \cdot \frac{1\text{ mol HA}}{1\text{ mol OH}^-}\\ \\ & = 0.00188\text{ mol HA}\end{aligned}`

Therefore, the molarity of the original solution was:

`\displaystyle \left[ \text{HA}\right] = \frac{0.00188\text{ mol}}{20.00\text{ mL}} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 0.0940\text{ M}`

In conclusion, the molarity of the unknown acid is about 0.0940 M.