Question

Chem 401 Discussion Workshop # 7 Name Exercise 12 NA C03 ( OHHcos For a 0,00345 M solution of sodjum carbonate: CNaO4CCO a) Write the Kh reaction: b) Determine the value of Kb (H,CO;: Ka 4.3 x 107, Kg- 5.6 x 10l) e) Calculate the [(OH-] by... cl) set up the ICE table c2) solve using the simplification approximation c3) Check approximation: c4) Solve using the quadratic d) Determine the pH of the solution. D7-6

Answer 1

Following is the - complete Answer -&- Explanation : for the given Question, in... Typed & Image Format...

$\Rightarrow$Answer(s):

1. Part - (a): Included in Part - (a), as discussed below...
2. Part - (b): The value of K_b = 1.786 x 10^-4( approx.)
3. Part - (c): [OH^-] = 0.000700 M (mole/L)
4. Part - (d): pH = 10.85

$\Rightarrow$Explanation:

Following is the complete Explanation for the above Answer(s) , in....Typed Format...

• Given:

1. ​​​​​​​Molar concentration ( Molarity ) of sodium carbonate (Na₂CO₃ ) = M_salt = 0.00345 M (mole/L)
2. For H₂CO₃ , K_a,1 = 4.3 x 10^-7 ; and K_a,2 = 5.6 x 10^-11

​​​​​​​   $\Rightarrow$Part - (a):

• Step - 1:

​​​​​​​We know Na₂CO₃, gets dissociated into ions in its aqueous solution, as follows, and the CO₂, obtained from the reaction, gets dissolved in water to produce, HCO₃^- and OH^-  ...These reactions are as follows

1. Na₂CO₃$\rightleftharpoons$ 2 Na⁺ + CO₃^2-   ----------------------------------- Equation - 1
2. CO₃^2- (aq) + H₂O (l)   $\rightleftharpoons$ HCO₃^- (aq)   + OH^- (aq) ----------- Equation - 2

Now, since Equation - 2, can be used to determine the value of K_b, for Na₂CO₃, we can call Equation - 2, as the K_b, reaction...

• Answer: Following is the K_b, reaction for Na₂CO₃( Answer - (a) )

​​​​​​​ CO₃^2- (aq) + H₂O (l)   $\rightleftharpoons$ HCO₃^- (aq)   + OH^- (aq)

$\Rightarrow$Part - (b):

• Step - 1:

We are given with the following:

1. K_a,1 =   4.3 x 10^-7
2. 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 H₂CO₃ ... at the above conditions...

• Step - 2:

​​​​​​​We know,for H₂CO₃, --- 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 [ HCO₃^- ] / [ CO₃ ^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.)

​​​​​​​$\Rightarrow$Part - (C):

• Step - 1:

​​​​​​​We will create our ICE Table, based on the above Equation - 2.... ( as below)

  CO₃^2- (aq) + H₂O (l)   $\rightleftharpoons$ HCO₃^- (aq)   + OH^- (aq) ----------- Equation - 2

• Step - 2:

​​​​​​​We know the initial concentration of CO₃^2- : [ CO₃^2-]_init = 0.00345 M ( mole/L)

Therefore, the ICE Table, will be :

	[ CO32-]	[HCO3-]	[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 [HCO₃^-] /  [ CO₃^2-] = X² / (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)

$\Rightarrow$Part - (d):

• Step - 1:

​​​​​​​We know:

pOH = - log₁₀[OH^-] = 3.1549

Since, pH + pOH = 14.....

Therefore:   pH = 14.0 - 3.15 = 10.85

• Answer:( part - (d) ):  

​​​​​​​ pH = 10.85 ( Answer )