Chapter 16 Principles of Chemical Reactivity: The Chemistry of Acids and Bases - 16-7 Calculations with Equilibrium Constants - Review & Check for Section 16-7 - Page 615: 3

B

$HCO_2^-(aq)+H_2O(l)\rightarrow HCO_2H(aq)+OH^-(aq)$ $K_b=[HCO_2H][OH^-]/[HCO_2^-]$ $[Na^+]=0.10\ M, [HCO_2H]=[OH^-]=x, [HCO_2^-]=0.10-x$ $5.6\times10^{-11}=x\dot{}x/(0.10-x)$ $x=2.37\times10^{-6}\ M$ $pH=14+log([OH^-])=8.37$ $[HCO_2^-]=0.10\ M$