Calculate the pH of 0.34 M ZnCl2 solution.
Express your answer using two decimal places.
ZnCl2 is a salt and it dissociate as:
ZnCl2 --> Zn2+ + 2Cl-
In aqueous solution we have
Zn2+ + 6H2O ----> [Zn(H2O)6]2+
It dissociate as:
[Zn(H2O)6]2+ <=> [Zn(H2O)5OH]+ + H+
From literature, Ka for [Zn(H2O)6]2+ is 2.5 *10-10
[Zn(H2O)6]2+ | <=> | [Zn(H2O)5OH]+ | H+ | |
I(M) | 0.34 | 0 | 0 | |
C(M) | -x | +x | +x | |
E(M) | 0.34-x | x | x |
Ka = [H+][Zn(H2O)5OH]+ / [Zn(H2O)6]2+
2.5 *10-10 = x2 /(0.34-x)
x2 + (2.5 *10-10 )x - (8.5 *10-11) = 0
x = 9.22 *10-6 = [H+]
pH = -log[H+] = - log (9.22 *10-6 ) = 5.035
Calculate the pH of 0.34 M ZnCl2 solution. Express your answer using two decimal places.
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