Determine [Zn2+] [CN-] and [HCN] in a saturated solution of Zn(CN)2 with a fixed ph of 2.900. The Ksp for Zn(CN)2 is 3.0*10^-16. The Ka for HCN is 6.2*10^-9
Zn(Cn)2 ------------> Zn+2 + 2
CN-
Ksp = [Zn+2 ] [ CN-
]2
let the conc of Zn+2 be s and conc of CN- be
2s
HCN ----------------> H+ +CN-
Ka = [H+] [CN-] / [HCN]
6.2*10-9 = [H+] [CN-] /
[HCN]
If PH = 2.9 then PH =
-log10[H+]
[H+] = 10-2.9 = 1.258925*10-3
M
1.258925*10-3 * [CN-] / [HCN] =
6.2*10-9
[HCN] =203048.3871 [CN-]
[Zn+2] = s ,
[CN-] + [HCN-] = 2s
[CN-] + 203048.3871 [CN-] = 2s
[CN-] = 9.84982*10-6 s
Ksp = [Zn+2 ] [ CN-
]2 , solve for s
3*10-16 = [s] * [ 9.84982*10-6 s
]2
3.092179*10-6 = s3
s = 0.0145687 M
Therefore [Zn+2] = 0.01456M ; [CN-] =2s =
0.029137M ;
Get Answers For Free
Most questions answered within 1 hours.