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sergey [27]
2 years ago
13

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?
Chemistry
1 answer:
ohaa [14]2 years ago
3 0

Answer:

About 0.0940 M.

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.

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