Substitute the concentrations into the equilibrium constant equation Ka-HIACVHAc H]=Ka×[ HACAC] =1.70×10-5×0.10/0.10 =1.70×105(mol.L) pH=477Substitute the concentrations into the equilibrium￾constant equation Ka = [H+ ][Ac- ]/[HAc] [H+ ] = Ka  [HAc]/[Ac- ] = 1.70  10-5  0.10/ 0.10 = 1.70  10-5 (mol.L-1 ) pH = 4.77