Answers
Following is the - complete Answer -&- Explanation : for the given Question, in... Typed & Image Format...
Answer(s):
- Part - (a): Included in Part - (a), as discussed below...
- Part - (b): The value of K_{b} = 1.786 x 10^{-4}( approx.)
- Part - (c): [OH^{-}] = 0.000700 M (mole/L)
- Part - (d): pH = 10.85
Explanation:
Following is the complete Explanation for the above Answer(s) , in....Typed Format...
- Given:
- Molar concentration ( Molarity ) of sodium carbonate (Na_{2}CO_{3} ) = M_{salt} = 0.00345 M (mole/L)
- For H_{2}CO_{3} , K_{a,1} = 4.3 x 10^{-7} ; and K_{a,2} = 5.6 x 10^{-11}
Part - (a):
- Step - 1:
We know Na_{2}CO_{3}, gets dissociated into ions in its aqueous solution, as follows, and the CO_{2}, obtained from the reaction, gets dissolved in water to produce, HCO_{3}^{-} and OH^{-} ...These reactions are as follows
- Na_{2}CO_{3} 2 Na^{+} + CO_{3}^{2-} ----------------------------------- Equation - 1
- CO_{3}^{2-} (aq) + H_{2}O (l) HCO_{3}^{-} (aq) + OH^{-} (aq) ----------- Equation - 2
Now, since Equation - 2, can be used to determine the value of K_{b}, for Na_{2}CO_{3}, we can call Equation - 2, as the K_{b}, reaction...
- Answer: Following is the K_{b}, reaction for Na_{2}CO_{3}( Answer - (a) )
CO_{3}^{2-} (aq) + H_{2}O (l) HCO_{3}^{-} (aq) + OH^{-} (aq)
Part - (b):
- Step - 1:
We are given with the following:
- K_{a,1} = 4.3 x 10^{-7}
- K_{a,2} = 5.6 x 10^{-11}
We should consider the value of K_{a,2} , as the value of acid dissociation constant for H_{2}CO_{3} ... at the above conditions...
- Step - 2:
We know,for H_{2}CO_{3}, --- K_{a,2} = 5.6 x 10^{-11} , and --- K_{w} = ( K_{a,2}) x ( K_{b} ) = 1.0 x 10^{-14}
And we know from Equation - 2:
K_{b} = [ OH^{-} ] x [ HCO_{3}^{-} ] / [ CO_{3} ^{2-} ] = K_{w} / K_{a,2} = ( 1.0 x 10^{-14} ) / ( 5.6 x 10^{-11} ) = 1.786 x 10^{-4}( approx.)
Therefore:
- Answer(b): The value of K_{b} = 1.786 x 10^{-4}( approx.)
Part - (C):
- Step - 1:
We will create our ICE Table, based on the above Equation - 2.... ( as below)
CO_{3}^{2-} (aq) + H_{2}O (l) HCO_{3}^{-} (aq) + OH^{-} (aq) ----------- Equation - 2
- Step - 2:
We know the initial concentration of CO_{3}^{2-} : [ CO_{3}^{2-}]_{init} = 0.00345 M ( mole/L)
Therefore, the ICE Table, will be :
[ CO_{3}^{2-}] | [HCO_{3}^{-}] | [OH^{-}] | |
Initial | 0.00345 M | 0.0 | 0.0 |
Change | - X | + X | + X |
Equilibrium | (0.00345 - X ) | + X | +X |
- Step - 3:
We know:
K_{b} = [OH^{-} ] x [HCO_{3}^{-}] / [ CO_{3}^{2-}] = X^{2} / (0.00345 - X ) = 1.786 x 10^{-4 }
Therefore: X = [OH^{-}] = 0.000700 M (mole/L) , i.e. solved using Quadratic Equation...
- Answer:
[OH^{-}] = 0.000700 M (mole/L)
Part - (d):
- Step - 1:
We know:
pOH = - log_{10}[OH^{-}] = 3.1549
Since, pH + pOH = 14.....
Therefore: pH = 14.0 - 3.15 = 10.85
- Answer:( part - (d) ):
pH = 10.85 ( Answer )