Question

use the references to access important values if needed for this question. the hydroxide ion concentration, oh⁻, of an aqueous solution of 0.501 m dimethylamine (a weak base with the formula (ch₃)₂nh), ( k_b = 5.9 \times 10^{-4} ), is oh⁻ = (\boxed{}) m.
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Explanation:

Step1: Define base dissociation equation

Dimethylamine dissociates as:

$$(\text{CH}_3)_2\text{NH} + \text{H}_2\text{O} ightleftharpoons (\text{CH}_3)_2\text{NH}_2^+ + \text{OH}^-$$

Step2: Set up $K_b$ expression

Let $x=[\text{OH}^-]=[(\text{CH}_3)_2\text{NH}_2^+]$, $[(\text{CH}_3)_2\text{NH}]=0.501-x$.
$$K_b = \frac{[(\text{CH}_3)_2\text{NH}_2^+][\text{OH}^-]}{[(\text{CH}_3)_2\text{NH}]} = \frac{x^2}{0.501-x}$$

Step3: Approximate $0.501-x\approx0.501$

Since $K_b$ is small, $x\ll0.501$, so:
$$5.9\times10^{-4} = \frac{x^2}{0.501}$$

Step4: Solve for $x$

$$x^2 = 5.9\times10^{-4} \times 0.501$$
$$x = \sqrt{5.9\times10^{-4} \times 0.501}$$
$$x = \sqrt{2.9559\times10^{-4}}$$
$$x \approx 0.0172$$

Answer:

$0.0172$ M