Calculate the pH of a solution obtained by mixing 100 mlof strong acid solution of pH=3 and 400ml solution of same acid solution of pH=1

millimomes of (acid)₁ = [H⁺]₁*100

millimomes of (acid)₂ = [H⁺]₂*400

net millimoles = 100( [H⁺]₁+4[H⁺]₂)

[H⁺]₁=10^-3      &    [H⁺]₂ = 10^-1

net millimoles = 100( 10^-3+4*10^-1) = (0.401)*100

concentration of resulting solution = total millimomes/total volume(in ml) = 0.401*100/500 = 8*10^-2 = [H⁺] (resulting sol)

PH =-log[H⁺]=-log8*10^-2 =2-log8 = 1.097