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?

Answer:

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

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\rightleftharpoons 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})