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 \([OH^-]\)

\([OH^-]=10^{-pOH}\)
\([OH^-]=10^{- 1.90}\approx1.26\times10^{-2}\space M\)

Step3: Set up the \(K_b\) expression

For a weak base \(B\): \(B + H_2O
ightleftharpoons BH^++OH^-\), \(K_b=\frac{[BH^+][OH^-]}{[B]}\)
Assume \(x = [OH^-]=[BH^+]\) (from the base - ionization reaction), and \([B]_{initial}=0.0910\space M\), \([B]=[B]_{initial}-x\). Since \(x = 1.26\times10^{-2}\space M\) and \([B]_{initial}=0.0910\space M\), \([B]=0.0910 - 1.26\times10^{-2}=0.0784\space M\)

Step4: Calculate \(K_b\)

\(K_b=\frac{x\times x}{[B]}\)
Substitute \(x = 1.26\times10^{-2}\space M\) and \([B]=0.0784\space M\)
\(K_b=\frac{(1.26\times10^{-2})^2}{0.0784}\)
\(K_b=\frac{1.59\times10^{-4}}{0.0784}\approx1.3\times10^{-3}\)

Answer:

\(K_b = 1.3\times10^{-3}\)