Question

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

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

Answer 1

`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))^(+)`

Answer 2

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