HNO3=========> H+ (aq) + NO3-(aq) [H+] = 0.0578 M [H+][OH-] =1.0*10^- 14 [OH-] = (1.0*10^-14)/[H+]= (1.0*10^-14)/0.0578 = 1.73*10^-13 pH = -log[H+] = - log(0.0578)=1.238=1.24 pH+pOH=14 pOH=14-pH=14-1.24=12.76 Solution HNO3=========> H+ (aq) + NO3-(aq) [H+] = 0.0578 M [H+][OH-] =1.0*10^- 14 [OH-] = (1.0*10^-14)/[H+]= (1.0*10^-14)/0.0578 = 1.73*10^-13 pH = -log[H+] = - log(0.0578)=1.238=1.24 pH+pOH=14 pOH=14-pH=14-1.24=12.76.