Both are going to use the same basic equation: Ka=[A-][H+]/[HA] We can replace the A- and H+ with x sincethey will be be the same concentration and HA with theconcentration of the acid 1. 1x10-5=[x][x]/.1 x=.001 pH=-log(x) =3 2. .01=[x][x]/.01 x=.01 pH=-log(x) pH=2 Solution Both are going to use the same basic equation: Ka=[A-][H+]/[HA] We can replace the A- and H+ with x sincethey will be be the same concentration and HA with theconcentration of the acid 1. 1x10-5=[x][x]/.1 x=.001 pH=-log(x) =3 2. .01=[x][x]/.01 x=.01 pH=-log(x) pH=2.