Question

an unknown weak base with a concentration of 0.0910 m has a ph of 12.10. what is the ( k_{b} ) of this base?

Explanation:

Step1: Calculate \(pOH\)

\(pH + pOH=14\), so \(pOH = 14 - pH\)
\(pOH=14 - 12.10 = 1.90\)

Step2: Calculate \([OH^-]\)

\([OH^-]=10^{-pOH}\)
\([OH^-]=10^{- 1.90}=1.26\times10^{-2}\space M\)

Step3: Set up the \(K_b\) expression

For a weak base \(B\) with reaction \(B + H_2O
ightleftharpoons BH^++OH^-\), \(K_b=\frac{[BH^+][OH^-]}{[B]}\)
Assume \(x = [OH^-]=[BH^+]\) (from the base - dissociation reaction), and \([B]=0.0910 - x\). Since \(x\) is small (\(0.0910\gg x\)), \([B]\approx0.0910\space M\)

Step4: Calculate \(K_b\)

\(K_b=\frac{(1.26\times10^{-2})(1.26\times10^{-2})}{0.0910}\)
\(K_b=\frac{1.59\times10^{-4}}{0.0910}=1.75\times10^{-3}\approx1.2\times10^{-3}\) (rounded for significant figures, considering the given data \(pH = 12.10\) has two decimal places)

Answer:

\(K_b = 1.2\times10^{-3}\)