Question

The concentration of $ {H^ + } $ and $ O{H^ - } $ in a $ 0.1M $ aqueous solution of $ 2\% $ ionized weak monobasic acid is [ ionic product of water = $ {10^{ - 14}} $ ]
A. $ 0.02 \times {10^{ - 3}}M $ and $ 5 \times {10^{ - 11}}M $
B. $ 1 \times {10^{ - 3}}M $ and $ 3 \times {10^{ - 11}}M $
C. $ 2 \times {10^{ - 3}}M $ and $ 5 \times {10^{ - 12}}M $
D. $ 3 \times {10^{ - 2}}M $ and $ 4 \times {10^{ - 13}}M $

Answer 1

Hint :This question can be solved by calculating the degree of dissociation for the dissociation of weak monobasic acid. Once a degree of dissociation is found, it will help us to find the value of concentration of hydronium ions. Once hydronium ion concentration is found we can calculate hydroxide ion by using the formula $ {K_W} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] $

Complete Step By Step Answer:
Let us firstly write down the reaction for this problem.
 $ HA\left( {aq} \right)\underset {} \leftrightarrows {H^ + }\left( {aq} \right) + {A^ - }\left( {aq} \right) $
It is given to us that $ \left[ {HA} \right] = 0.1M $ and it is dissociated $ 2\% $
Therefore, degree of dissociation will be $ \left( \alpha \right) = \dfrac{2}{{100}} \times 0.1 = 0.002 $
Now, from the above equation we see that 1 mole of acid dissociates and forms 1 mole of hydronium ion and anion.
Also, we know that degree of dissociation is the fraction of a mole of the reactant that underwent dissociation.
This means that from the original concentration which was $ \left[ {0.1} \right]M $ , $ \left[ {0.002} \right]M $ will be dissociated. And the same concentration $ \left[ {0.002} \right]M $ of $ {H^ + } $ is formed.
Hence, we get the concentration of $ [{H^ + }] = 0.002M $ or $ 2 \times {10^{ - 3}}M $
Now we know that the ionic product of water $ \left( {{K_W}} \right) $ is defined as the product of concentration of $ {H^ + }\& O{H^ - } $ ions at a particular temperature. This can be expressed as
 $ {K_W} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] $
Now, $ {K_W} $ for water is given in the question which is equal to $ {10^{ - 14}} $ . And we have already calculated $ \left[ {{H^ + }} \right] = 2 \times {10^{ - 3}}M $
Substituting these values in the above expression of ionic product we can find the concentration of hydroxide ions.
 $ {K_W} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] $
 $ \Rightarrow {10^{ - 14}} = \left[ {2 \times {{10}^{ - 3}}} \right]\left[ {O{H^ - }} \right] $
 $ \Rightarrow \left[ {O{H^ - }} \right] = \dfrac{{{{10}^{ - 14}}}}{{2 \times {{10}^{ - 3}}}} $
 $ \Rightarrow \left[ {OH} \right] = 5 \times {10^{ - 12}}M $
Therefore, concentration of [OH]= $ 5 \times {10^{ - 12}}M $
Hence, the correct option is C.

Note :
It should be noted that the ionic product of water $ {K_W} $ depends on the temperature of the water and its value changes with the change in temperature. This is the most common fact which is misunderstood. Students think that the value of an ionic product is always $ {10^{ - 14}} $ . But this is the value at standard temperature and its values will change with change in temperature.