Question #19507

Consider a solution that is 0.10 M in a weak triprotic acid which is represented by the gen- eral formula H3A with ionization constants K1 = 1.0×10−3, K2 = 1.0×10−8, and K3 = 1.0×10−12. What is the pH of the solution?

Expert's answer

K_{3} is very small,so K_{2} and K_{1} is resulting infinding pH

H_{2}A^{-} = H^{+} + HA^{-}

K_{2} = [H^{+}]^{2}/H_{2}A^{-}

H_{2}A^{-} = [H^{+}]^{2}/K_{2}

H_{3}A = H^{+} + [H^{+}]^{2}/K_{2}

K_{1} = ([H^{+}]^{2}/K_{2})^{2}/[H_{3}A]

[H^{+}]^{4} = K_{1}*K_{2}^{2}[H_{3}A]

[H^{+}] = (K_{1}*K_{2}^{2}[H_{3}A])^{1/4} = (1.0×10^{−3}*(1.0×10^{−8})^{2} * 0.10)^{1/4} = 1x10^{10}

pH = -log [H^{+}] = 10

H

K

H

H

K

[H

[H

pH = -log [H

## Comments

## Leave a comment