Question

Calculate the concentrations of all species in a 1.02 M Na.SO, (sodium sulfite) solution. The ionization constants for sulfurous acid are K1 = 1.4 x 10-2 and Ka = 6,3 x 10- (Nat] = M (S01 (HSO;] MH,80] = (OH) = M (H+) =

Answer 1

Ans :-

Na₂SO₃ (aq) -------------------> 2 Na⁺ (aq) + SO₃^2- (aq)

Given,

Concentration of Na₂SO₃ = [Na₂SO₃] = 1.02 M

[Na⁺] = 2 x 1.02 M = 2.04 M

[SO₃^2-] = 1.02 M

K_b1of SO₃^2- = K_w/K_a2 = 1.0 x 10^-14 / 6.3 x 10^-8 = 1.59 x 10^-7

ICE table of SO₃^2- :

............................SO₃^2- (aq)............+.............H₂O (l) <-------------------> HSO₃^- (aq)........+..........OH^- (aq)

Initial...................1.02 M...........................................................................0.00 M............................0.00 M

Change................-y..................................................................................+y.....................................+y

Equilibrium...........(1.02-y) M........................................................................y M....................................y M

Expression of K_b1 is :

K_b1 = [HSO₃^- ].[OH^-] / [SO₃^2-]

1.59 x 10^-7 = y²/(1.02-y)

If y <<<1.02 , then neglect y

1.59 x 10^-7 = y²/(1.02)

1.62 x 10^-7 = y²

y = 4.02 x 10^-4 M

So, [HSO₃^- ] = [ OH^-] = 4.02 x 10^-4 M

Also,

[H⁺] = 1.0 x 10^-14 / 4.02 x 10^-4​​​​​​​  M

[H⁺] = 2.5 x 10^-11 M

K_b2of HSO₃^- = K_w/K_a1 = 1.0 x 10^-14 / 1.4 x 10^-2 = 7.14 x 10^-13

ICE table of SO₃^2- :

............................HSO₃^- (aq)............+.............H₂O (l) <-------------------> H₂SO₃ (aq)........+..........OH^- (aq)

Initial...................4.02 x 10^-4​​​​​​​ M ...............................................................0.00 M.........................4.02 x 10^-4​​​​​​​  M

Change................-y..................................................................................+y.....................................+y

Equilibrium...........(4.02 x 10^-4​​​​​​​ M-y) M..................................................y M..............................(4.02 x 10^-4​​​​​​​ +y) M

Expression of K_b1 is :

K_b2 = [H₂SO₃ ].[OH^-] / [HSO₃^-]

7.14 x 10^-13 = y((4.02 x 10^-4​​​​​​​ +y) /(4.02 x 10^-4​​​​​​​ -y)

If y is very small , then neglect y

7.14 x 10^-13 = y((4.02 x 10^-4​​​​​​​ ) /(4.02 x 10^-4​​​​​​​ )

y =7.14 x 10^-13 M

So, [H₂SO₃ ] = 7.14 x 10^-13 M​​​​​​​

Hence,

[Na⁺] = 2.04 M

[SO₃^2-] = 1.02 M

[H⁺] = 2.5 x 10^-11 M

[ OH^-] = 4.02 x 10^-4​​​​​​​ M

[HSO₃^- ] = 4.02 x 10^-4​​​​​​​ M

[H₂SO₃ ] = 7.14 x 10^-13 M​​​​​​​