## Chemistry and Chemical Reactivity (9th Edition)

a) Initial concentration: $0.515\ g\div 94.11\ g/mol\div 125/1000\ L=0.044\ M$ $K_a=1.3\cdot10^{-10}=x\cdot x/(0.044\ M - x)$ $x=2.39\cdot10^{-6}\ M=[H_3O^+]$ $pH=5.62$ b) With all the hydroxide being consumed: The initial number of moles of hydroxide: $0.044\ M\times 125/1000=0.0055\ mol$ Volume added: $0.0055\ mol\div 0.123\ M=44.7\ mL$ The concentration of the phenol anion: $0.0055\ mol\div (125+44.7)/1000\ L=0.0324\ M$ Equilibrium: $K_w/K_a=[OH^-][Hp]/[p^-]=x\cdot x/(0.0323-x)$ $x=0.0015\ M=[OH^-], [C_6H_5O^-]=0.0308\ M$ $[H_3O^+]=6.7\cdot10^{-12}\ M$ Sodium is just diluted: $[Na^+]=0.123\ M\times 44.7/(125+44.7)=0.0323\ M$ c) $pH=11.18$
a) Initial concentration: $0.515\ g\div 94.11\ g/mol\div 125/1000\ L=0.044\ M$ $K_a=1.3\cdot10^{-10}=x\cdot x/(0.044\ M - x)$ $x=2.39\cdot10^{-6}\ M=[H_3O^+]$ $pH=5.62$ b) With all the hydroxide being consumed: The initial number of moles of hydroxide: $0.044\ M\times 125/1000=0.0055\ mol$ Volume added: $0.0055\ mol\div 0.123\ M=44.7\ mL$ The concentration of the phenol anion: $0.0055\ mol\div (125+44.7)/1000\ L=0.0324\ M$ Equilibrium: $K_w/K_a=[OH^-][Hp]/[p^-]=x\cdot x/(0.0323-x)$ $x=0.0015\ M=[OH^-], [C_6H_5O^-]=0.0308\ M$ $[H_3O^+]=6.7\cdot10^{-12}\ M$ Sodium is just diluted: $[Na^+]=0.123\ M\times 44.7/(125+44.7)=0.0323\ M$ c) $pH=11.18$