Question

If $Pb(s)|PbS{O_4}(s)|NaHS{O_4}(0.600M)||P{b^{2 + }}(2.50 \times {10^{ - 5}}M)|Pb(s)$
The voltage of the cell in $V$ is $E = + 0.061V$ . Calculate ${K_2} =$
$[{H^ + }][SO_4^{2 - }]/[HSO_4^ - ]$ , the dissociation constant for $HSO_4^ -$ _____.
Given $Pb(s) + SO_4^{2 - } \to PbS{O_4} + 2{e^ - }(E^\circ = 0.356V),E^\circ (P{b^{2 + }}/Pb) = - 0.126V.$
Multiply the answer by $100$ and fill in the blanks. (write the value to the nearest integer)

Answer 1

Hint : First calculate the value of $E{^\circ _{cell}}$ . The value of $E{^\circ _{cell}}$ will be used in the Nernst equation to get the value of $Q$ which is a ratio of molarity of products to molarity of reactants. The value of $Q$ will be used to calculate the molarity of $SO_4^{2 - }$ . Now we can calculate the dissociation constant $K$ .

Complete Step By Step Answer:
First we have to calculate the value of $E{^\circ _{cell}}$ ,
We know that,
$E{^\circ _{cell}} = E{^\circ _R} - E{^\circ _L}$
It is given that,
$E{^\circ _R} = - 0.126$ and $E{^\circ _L} = 0.356$
Therefore,
$E{^\circ _{cell}} = - 0.126 + 0.356$
$E{^\circ _{cell}} = 0.23V$
We can now use Nernst's equation to calculate the value of $Q$ . Nernst equation is given as:
${E_{cell}} = E{^\circ _{cell}} - \dfrac{{0.0592}}{n}\log Q$
Where, ${E_{cell}} \to$ max potential which can be generated when no current is flowing.
$E{^\circ _{cell}} \to$ cell potential
$n \to$ number of electrons gained or lost during reaction.
If we check the reaction, we can see that the number of electrons lost is $2$ .
$Pb(s) + SO_4^{2 - } \to PbS{O_4} + 2{e^ - }$
On putting the values in Nernst equation,
$0.061 = 0.23 - \dfrac{{0.0592}}{2}\log Q$
On further solving,
$\log Q = 5.709$
$Q = 5.122 \times {10^5}$
Since,
$Q = \dfrac{{[products]}}{{[reac\tan ts]}}$
$Q = \dfrac{{[PbS{O_4}]}}{{[P{b^{2 + }}][SO_4^{2 - }]}}$
Hence,
$5.122 \times {10^5} = \dfrac{1}{{2.5 \times {{10}^{ - 5}} \times [SO_4^{2 - }]}}$
$[SO_4^{2 - }] = \dfrac{1}{{5.122 \times {{10}^5} \times 2.5 \times {{10}^{ - 5}}}}$
$[SO_4^{2 - }] = 0.078M$
Hence, $0.078M$ of $SO_4^{2 - }$ will make ${E_{cell}} = 0$ , thus, we reach equilibrium and now we can calculate the value of dissociation constant.
The dissociation constant $K$ for $HSO_4^ -$ can be calculated as follows,
$K = \dfrac{{[{H^ + }][SO_4^{2 - }]}}{{[HSO_4^ - ]}}$
$K = \dfrac{{0.078 \times 0.078}}{{0.6}}$
Hence,
$K = 0.0106$
Since we are asked to multiply the value of $K$ by $100$ , thus, on multiplying,
$K \times 100 = 1.06$
On writing the value of $K$ to its nearest integer,
$\to K = 1.06 \approx 1$
Hence the value of $K$ is $1$ .

Note :
Neither the value of $Q$ nor the dissociation constant $K$ has any unit as both of them are simply a ratio of molarity of products to their reactants. $E{^\circ _R}$ and $E{^\circ _L}$ can be understood as potential at anode and potential at cathode respectively. The dissociation constant $K$ can also be written as ${K_d}$ .
The dissociation constant is called ionization constant when it is applied for salts. The inverse of dissociation constant is called association constant.