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An unknown weak base has a concentration of 0.170 M and a pH of 9.42. What is the value of K_b for this base?
A
7.5 × 10^{-3}
B
1.2 × 10^{-7}
C
2.6 × 10^{-5}
D
4.8 × 10^{-10}
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First, use the given pH to find the pOH. Recall that pH + pOH = 14, so calculate pOH as \(\text{pOH} = 14 - \text{pH}\).
Next, find the hydroxide ion concentration \([\mathrm{OH}^-]\) using the pOH value with the formula \([\mathrm{OH}^-] = 10^{-\text{pOH}}\).
Set up an expression for the base dissociation equilibrium. For a weak base B, the reaction is \(\mathrm{B} + \mathrm{H_2O} \rightleftharpoons \mathrm{BH}^+ + \mathrm{OH}^-\), and the initial concentration is given as 0.170 M.
Assuming \(x = [\mathrm{OH}^-]\) produced at equilibrium, write the expression for \(K_b\) as \(K_b = \frac{x^2}{[\mathrm{B}]_\text{initial} - x}\). Since \(x\) is small compared to the initial concentration, you can approximate the denominator as the initial concentration.
Finally, substitute the values of \(x\) and the initial concentration into the \(K_b\) expression to calculate the base dissociation constant.