Chapter 2: Q172CP (page 40)

A 1000m solution of chloroacetic acid $({ClCH}_{2} COOH)$ freezes at $1.93 ° C$ . Find the Ka of chloroacetic acid. (Assume the molarities equal the molalities.)

Short Answer

Expert verified

$K_{a} = 1.50 \times 10^{- 3} .$

Step by step solution

Find the Ka

Important things and equations to note:

$[{ClCH}_{2} COOH] = 1.000m = 1.000M$

Freezing point depression: $Δ T_{f} = {iK}_{f} m$

Degree of dissociation of a weak acid is:

$α = \frac{i - 1}{n - 1}$ , where n is the number of ions.

$K_{a} = \frac{[HA] α^{2}}{1 - α}$

Now, solve for the i using the freezing point depression formula. Note that for the original freezing point and $K_{f}$ , we will use these values from water since it depends on the solvent.

$Δ T_{f} = {iK}_{f} mi= \frac{Δ T_{f}}{K_{f} m} = \frac{0.00°C- (- 1.93°C)}{(1.86°C/m) (1.000m)} i = 1.038$

Now, replace the i in the degree of dissociation of a weak acid formula to solve $α$ . Note that ${ClCH}_{2} COOH$ will produce 2 ions.

${ClCH}_{2} COOH + H_{2} O ⇌ {ClCH}_{2} {COO}^{-} {+H}_{3} O^{+}$

$α = \frac{i - 1}{n - 1} = \frac{1.038 - 1}{2 - 1} α = 0.038$

Lastly, solve for the $K_{a}$ .

$\begin{array}{rcl} K_{a} = \frac{[HA] α^{2}}{1 - α} \\ = \frac{(1.000) ({0.038}^{2})}{1 - 0.038} \\ = 1.50 \times 10^{- 3} . \end{array}$

Hence, $K_{a} = 1.50 \times 10^{- 3} .$

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	Mass (g) of Sample	Mass(g) of Sulfur in Sample
Coal A	378	11.3
Coal B	495	19.0
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A 1000m solution of chloroacetic acid $({ClCH}_{2} COOH)$ freezes at $1.93 ° C$ . Find the Ka of chloroacetic acid. (Assume the molarities equal the molalities.)

Short Answer

Step by step solution

Find the Ka

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A 1000m solution of chloroacetic acid(ClCH2COOH)freezes at 1.93°C. Find the Ka of chloroacetic acid. (Assume the molarities equal the molalities.)

Short Answer

Step by step solution

Find the Ka

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Inorganic Chemistry

Kinetics

Physical Chemistry

Nuclear Chemistry

Chemical Analysis

Study anywhere. Anytime. Across all devices.

A 1000m solution of chloroacetic acid $({ClCH}_{2} COOH)$ freezes at $1.93 ° C$ . Find the Ka of chloroacetic acid. (Assume the molarities equal the molalities.)