Question

Using the systematic method and information in the Appendices F and J (note that log value is given in Appendix J), determine [SCN-], [Hg2+], and [HgSCN+] at equilibrium when Hg(SCN)2(s) is dissolved in water.

Answer 1

If you will look up those values and post them here we may be able to help. Include infor on HgSCN+. Why? My table values may not agree (probably will not) so we can't get the same answer. Show what you know to do also.

Answer 2

I posted the whole question again. Hg2+and SCN-=HgSCN+ =logarithim for the reaction is 9.64 = 10^9.64= 4.3*10^9

Ksp HG2(SCN)2= 3.0*10^-20
I have reactions Hgscn2> hg2+2SCN
Hg2+ + SCN->Hgscn
cations= anions
2[hg2+]+[Hgscn]=[Scn-]
Mass balance
[hgscn+]+[Hg2+]=[HGscn+]+[Scn-]