# What is K_a of "ZnCl"_2 ?

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

Mar 12, 2017

$2.5 \cdot {10}^{- 10}$

#### Explanation:

I was able to find a value for the acid dissociation constant of the ${\text{Zn}}^{2 +}$ cations here

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

Of course, seeing how ${\text{ZnCl}}_{2}$ is a soluble ionic compound and knowing that the chloride anion does not hydrolyze in water, the acid dissociation constant actually belongs to the ${\text{Zn}}^{2 +}$ cations.

You will have

${\text{ZnCl"_ (2(aq)) -> "Zn"_ ((aq))^(2+) + 2"Cl}}_{\left(a q\right)}^{-}$

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

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 (${H}^{+}$ = ${10}^{-} 4.4$)
$\sqrt{0.1 x} = {H}^{+} \mathmr{and} {10}^{-} 4.4$
or
$0.1 x = {0.000044}^{2}$

$0.1 x = 1.936E-9$

$x = K a = 1.936E-8 = \setminus \frac{1.936E-9}{0.1}$

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