## Chemistry and Chemical Reactivity (9th Edition)

a) The first because it has a higher Kb b) $K_b=[BH^+][OH^-]/[B]$ $4.3\cdot10^{-4}=x\cdot\ x/(0.10-x)$ $x=[OH^-]=0.0063\ M$ $pH=14+log\ [OH^-]=11.80$