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2 years ago

A pure liquid has a constant boiling point, but a liquid with a constant boiling point is not necessarily pure. Explain.

1 answer:

7 0

A pure liquid has a constant boiling point (at constant pressure). A change in boiling during distillation is an indication of impurity. The opposite, proposition, however, is not always true, and a constant boiling point does not necessarily mean that the liquid is pure.

This is because, at the boiling point, vapor and liquid are in equilibrium, and, if the composition of the vapor and liquid remains constant through the process, the temperature will also remain constant.

The boiling point can differ by location depending on pressure. Why does not a pure liquid in a distilling flask vaporize all at once whilst the boiling temperature is reached? Additional heat must be supplied for a phase change to occur.

Learn more about a boiling point here brainly.com/question/40140

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3 years ago

Calculate the pKa of lactic acid (CH3CH(OH)COOH) given the following information. 3.005 grams of potassium lactate are added to

snow_lady [41]

Answer:

$\displaystyle \text{p} K_a \approx 3.856$

Explanation:

Because 3.005 grams of potassium lactate is added to 100. mL of solution, its concentration is:

$\displaystyle \begin{aligned} \left[ \text{KC$_3$H_$_5$O$_3$}\right] & = \frac{3.005\text{ g KC$_3$H_$_5$O$_3$}}{100.\text{ mL}} \cdot \frac{1\text{ mol KC$_3$H_$_5$O$_3$}}{128.17 \text{ g KC$_3$H_$_5$O$_3$}} \cdot \frac{1000\text{ mL}}{1\text{ L}} \\ \\ &= 0.234\text{ M}\end{aligned}$

By solubility rules, potassium is completely soluble, so the compound will dissociate completely into potassium and lactate ions. Therefore, [KC₃H₅O₃] = [C₃H₅O₃⁺]. Note that lactate is the conjugate base of lactic acid.

Recall the Henderson-Hasselbalch equation:

$\displaystyle \begin{aligned}\text{pH} = \text{p}K_a + \log \frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]} \end{aligned}$

[Base] = 0.234 M and [Acid] = 0.500 M. We are given that the resulting pH is 3.526. Substitute and solve for p<em>Kₐ</em>:

$\displaystyle \begin{aligned} (3.526) & = \text{p}K_a + \log \frac{(0.234)}{(0.500)} \\ \\ 3.526 & = \text{p}K_a + (-0.330) \\ \\ \text{p}K_a & = 3.856\end{aligned}$

In conclusion, the p<em>Kₐ </em>value of lactic acid is about 3.856.

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