Question

If osmotic pressure of 1 M aqueous solution of $ {H_2}S{O_4} $ at 500 K is 90.2 atm. Calculate $ {K_a}_{_2} $ of $ {H_2}S{O_4} $ . Give your answer after multiplying 1000 with $ {K_a}_{_2} $ . (Assume ideal solution) [Given: $ {K_a}_{_1}{\text{ }}of{\text{ }}{H_2}S{O_4} = \infty ,R = 0.082L{\text{ }}atm{\text{ }}mo{l^{ - 1}}{K^{ - 1}} $ ]

Answer 1

Hint: We know for ideal solution, we can write:
 $ \pi = iCRT $
here,
 $ \pi = osmotic{\text{ }}pressure $
 $ C = {\text{ }}concentration{\text{ }}of{\text{ }}solution $
 $ R = {\text{ }}Universal{\text{ }}Gas{\text{ }}constant $
 $ T = {\text{ }}Temperature $
To calculate it, we need dissociation of $ {H_2}S{O_4} $ for 2 times and calculate the value of $ {K_a}_{_2} $ using the dissociation equation.

Complete step by step solution:
Firstly we need to write first dissociation of $ {H_2}S{O_4} $ as given below:
 $ {H_2}S{O_4} \to {H^ + } + HS{O_4}^ - $
Since the first dissociation constant $ {K_a}_{_1} = \infty $ , that means complete dissociation will happen, so we can write:
 $ {\text{ }}{H_2}S{O_4} \to {H^ + } + HS{O_4}^ - \\
  {\text{initial: 1 0 0}} \\
  {\text{final: 0 1 1}} \\
 $
Thus, $ {H^ + }{\text{ and H}}S{O_4}^ - $ both have a concentration of 1M.
Now let us find out second dissociation constant $ {K_a}_{_2} $ of acid as below equation:
 $ HS{O_4}^ - \rightleftharpoons {H^ + } + S{O_4}^{2 - } $
Now, let us assume that acid does not dissociate completely, so we assume that $ x $ concentration is dissociated, so using stoichiometry, we can say that ‘x’ is concentration of $ {H^ + }{\text{ and }}S{O_4}^{2 - } $
We also know that $ {H^ + } $ is a common ion and from first dissociation it has concentration of 1M, so that is the initial concentration of $ {H^ + } $ even in this reaction.
 $
  {\text{ }}HS{O_4}^ - \rightleftharpoons {H^ + } + S{O_4}^{2 - } \\
  {\text{initial: 1 1 0}} \\
  {\text{final: 1 - x 1 + x x}} \\
$
Now, we can write i, Van’t Hoff factor, which can be written as total ions dissociated.
 $ i = 1 - x + 1 + x + x = 2 + x $
Given data in question is:
 $ \pi = {\text{osmotic pressure}} = 90.2atm \\
  C = 1M \\
  R = 0.082L{\text{ }}atm{\text{ mo}}{{\text{l}}^{{\text{ - 1}}}}{{\text{K}}^{{\text{ - 1}}}} \\
  T = 500K \\
 $
Now we know for ideal solution, the equation for osmotic pressure is given by:
 $ \pi = iCRT $
Now we substitute all values in the above equation, so we get:
 $ 90.2 = (2 + x) \times 1 \times 0.082 \times 500 $
Simplifying, and keeping 2+x on one side we can calculate the values as:
 $\ \dfrac{{90.2}}{{1 \times 0.082 \times 500}} = 2 + x \\
   \Rightarrow \dfrac{{90.2}}{{41}} = 2 + x \\
 $
On simplification, we can write:
 $ \Rightarrow 2.2 = 2 + x $
We keep x on one side, we can simplify:
 $ \Rightarrow x = 2.2 - 2 = 0.2 $
Thus, the concentration of ions by substituting the value of ‘x’ can be written as:
 $ [{H^ + }] = 1 + x \\
   = 1 + 0.2 \\
   = 1.2 \\
$
 $ [S{O_4}^{2 - }] = x \\
   = 0.2 \\
$
 $ [HS{O_4}^ - ] = 1 - x \\
   = 1 - 0.2 \\
   = 0.8 \\
 $
Now, to calculate $ {K_a}_{_2} $ , we can write:
 $ {K_a}_{_2} = \dfrac{{[{H^ + }][S{O_4}^{2 - }]}}{{[HS{O_4}^ - ]}} $
Now, substitute the concentration of all ions and simplify the equation:
 $ {K_a}_{_2} = \dfrac{{1.2 \times 0.2}}{{0.8}} = 0.3 $
Thus, we can multiply 1000 by the above value to get 300 as an answer.
So, the final answer is 300.

Note:
When dissociation constant value is infinity, then the dissociation is complete as in case of first dissociation of acid.
In substitute for the equation of osmotic pressure for an ideal solution, we may forget the ‘i’, Van’t Hoff factor. So calculate carefully and substitute.
The problem is lengthy, so solve it carefully and step by step to reach the final answer, and at last do not forget to multiply by 1000, to get the final answer.