Zn(CN)2 Zn2+ + 2CN-
Ksp = [Zn2+] [CN-]2 = 3*10-16
Considering the solubility product to be s , the above equation can be written as
s*s2= 3*10-16
Thus , s = cube root of 3*10-16 = 6.69 *10-6
[Zn2+] = s= 6.69 *10-6 and [CN-]= square root of 6.69 *10-6 = 2.58 * 10-3
pH =pKa + log [salt]/[acid]
2.180 = -9.2 + log {(3*10-16) / [acid]}
11.38 = log (3*10-16) - log [acid]
(-26.9) = log [acid] [acid] = antilog (-26.9) = 1.25 * 10-27
LL Determine [Zn2+1, (CN"), and [HCN) in a saturated solution of Zn(CN), with a fixed pH...
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