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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
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
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