What is #K_a# of #"ZnCl"_2# ?

I know its small but what is the exact . Tell me the source too

2 Answers
Mar 12, 2017

Answer:

#2.5 * 10^(-10)#

Explanation:

I was able to find a value for the acid dissociation constant of the #"Zn"^(2+)# cations here

http://bilbo.chm.uri.edu/CHM112/tables/KaTable.htm

Of course, seeing how #"ZnCl"_2# is a soluble ionic compound and knowing that the chloride anion does not hydrolyze in water, the acid dissociation constant actually belongs to the #"Zn"^(2+)# cations.

You will have

#"ZnCl"_ (2(aq)) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)#

followed by

#"Zn"_ ((aq))^(2+) + 6"H"_ 2"O"_ ((l)) -> ["Zn"("H"_ 2"O")_ 6]_ ((aq))^(2+)#

and

#["Zn"("H"_ 2"O")_ 6]_ ((aq))^(2+) + "H"_ 2"O"_ ((l)) rightleftharpoons ["Zn"("H"_ 2"O")_ 5("OH")]_ ((aq))^(+) + "H"_ 3"O"_ ((aq))^(+)#

Mar 12, 2017

Answer:

The Ka is nearly #1.936E-8#

Explanation:

http://www.webassign.net/labsgraceperiod/ucscgencheml1/lab_8/manual.pdf

In this website I checked that pH of 0.1M of ZnCl2 is 4.4
I tried to solve for Ka (#H^+# = #10^-4.4#)
#sqrt(0.1x) = H^+ or 10^-4.4#
or
#0.1x = 0.000044^2#

#0.1x= 1.936E-9#

#x = Ka = 1.936E-8 = \frac{1.936E-9}{0.1}#

You can solve this by ICE table. Answer would be the same

The Ka is #1.936E-8..# and If you check in this website what is the Ka of Zn(H2O)6 it would be the same . Just they have rounded it off to simpler numbers. These dont matter much