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.