Question #133500

To find volume of water, which should be added to 11 of a 0,6 % solution of an acetic acid (p = 1) for reception of a solution with pH =3 (Kd = 1,8 10).

Expert's answer

The concentration of the solution is unknown.

The pH of the solution is 3.003.00

The new volume is unknown.

The equilibrium equation is;

CH3COOH⇋H++CH3COO−CH_3COOH\leftrightharpoons H^++CH_3COO^- At equilibrium, Co=1.8×10−5C_o=1.8\times 10^-5

Kd=K_d=[H+][CH3COO−][CH3COOH][H^+][CH_3COO^-]\over[CH_3COOH]

== [1.8×10−5]×[1.8×10−5][−1.8×10−5][1.8\times 10^{-5}]\times[1.8\times 10^{-5}]\over [-1.8\times 10^{-5}] =3.6×10−5=3.6\times 10^{-5}

[H+]=10−3.00[H^+]=10^{-3.00}

The volume in litres of the solution is;

V=V= 13.6×10−51\over 3.6\times 10^{-5} =2.7×104=2.7\times 10^4 LL

T

0.61000.6\over 100 ×2.7×104=1.6×102L\times 2.7\times 10^4=1.6\times 10^2L


The new volume == 1.6×102L1.6\times 10^2 L


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