Multiple Choice
Calculate the pH of a solution that is 0.205 M in CH3NH2 and 1.55 M in CH3NH3Br. Assume the Kb for CH3NH2 is 4.4 x 10^-4.
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17. Acid and Base Equilibrium
pH of Weak Bases
Multiple Choice
What is the pH of a 0.280 M solution of (CH_3)_2NH (dimethylamine)? The K_b of (CH_3)_2NH is 5.4 \(\times\) 10^{-4}.
A
12.10
B
2.90
C
11.23
D
9.85
Verified step by step guidance1
Identify the species involved: (CH_3)_2NH (dimethylamine) is a weak base. We are given its concentration (0.280 M) and its base dissociation constant, K_b = 5.4 \(\times\) 10^{-4}.
Write the base dissociation equilibrium for dimethylamine in water: \(\mathrm{(CH_3)_2NH + H_2O \rightleftharpoons (CH_3)_2NH_2^+ + OH^-}\).
Set up an ICE table (Initial, Change, Equilibrium) to express the concentrations of species at equilibrium. Let \(x\) be the concentration of OH\(^-\) produced:
\[\begin{array}{c|ccc} & \mathrm{(CH_3)_2NH} & \mathrm{OH^-} & \mathrm{(CH_3)_2NH_2^+} \\ \hline \text{Initial (M)} & 0.280 & 0 & 0 \\ \text{Change (M)} & -x & +x & +x \\ \text{Equilibrium (M)} & 0.280 - x & x & x \end{array}\]
Write the expression for the base dissociation constant \(K_b\) in terms of \(x\(: \[K_b = \frac{[\mathrm{OH^-}][(CH_3)_2NH_2^+]}{[(CH_3)_2NH]} = \frac{x \times x}{0.280 - x} \approx \frac{x^2}{0.280}\] (assuming \)x\) is small compared to 0.280).
Solve for \(x\) (the OH\(^-\) concentration) by rearranging the equation: \[x = \sqrt{K_b \times 0.280}.\] Then calculate the pOH using \[\mathrm{pOH} = -\log[OH^-] = -\log x,\] and finally find the pH using \[\mathrm{pH} = 14 - \mathrm{pOH}.\]
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