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:

$K_a=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

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

$1.3\times 10^{-2}=\frac{(x)\times (x)}{(0.35-x)}$

$x=0.061M$

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