Question

What are the concentrations of hso4−, so42-, and h+ in a 0.35 m khso4 solution? (hint: h2so4 is a strong acid; ka for hso4− = 1.3 ✕ 10−2.)?

Answer 1

Answer;

[H+] = 0.051 M

[SO4=] = 0.051M

[HSO4-] = 0.16M

Explanation;

HSO4- <==> H+ + SO4= ..... Ka = 1.3x10^-2

Ka = [H+] [SO4=] / [HSO4-]

1.3x10^-2 = x^2 / (0.21-x)

Using algebra and solving the quadratic equation to solve for x.

x = 0.051

Therefore;

[H+] = [SO4=] = 0.051M

[HSO4-] = 0.16M

Answer 2

Answer : The concentration of
`HSO_4^-`,
`SO_4^(2-)` and
`H^+` are 0.29 M, 0.061 M and 0.061 M respectively.

Explanation :

First we have to calculate the concentration of
`HSO_4^-`

The dissociation of
`KHSO_4` is:

`KHSO_4\rightarrow K^++HSO_4^-`

As, 1 mole of
`KHSO_4` gives 1 mole of
`HSO_4^-`

So, 0.35 M of
`KHSO_4` gives 0.35 M of
`HSO_4^-`

Now we have to determine the concentration of
`SO_4^(2-)` and
`H^+`.

The dissociation of
`HSO_4^-` is:

`HSO_4^-\rightleftharpoons H^++SO_4^(2-)`

Initial conc. 0.35 0 0

At eqm. (0.35-x) x x

The expression of acid dissociation constant will be:

`K_a=([H^+][SO_4^(2-)])/([HSO_4^-])`

Now put all the given values in this expression, we get:

`1.3* 10^(-2)=((x)* (x))/((0.35-x))`

`x=0.061M`

Thus, the concentration of
`SO_4^(2-)` = x = 0.061 M

The concentration of
`H^+` = x = 0.061 M

The concentration of
`HSO_4^-` = 0.35 - x = 0.35 - 0.061 = 0.29 M