1 answer

Chem 401 Discussion Workshop # 7 Name Exercise 12 NA C03 ( OHHcos For a 0,00345...

Question:

Chem 401 Discussion Workshop # 7 Name Exercise 12 NA C03 ( OHHcos For a 0,00345 M solution of sodjum carbonate: CNaO4CCO a) W
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

Answers

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

\RightarrowAnswer(s):

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

\RightarrowExplanation:

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

  • Given:
  1. ​​​​​​​Molar concentration ( Molarity ) of sodium carbonate (Na2CO3 ) = Msalt = 0.00345 M (mole/L)
  2. For H2CO3 , Ka,1 = 4.3 x 10-7 ; and Ka,2 = 5.6 x 10-11

​​​​​​​   \RightarrowPart - (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+ + CO32-   ----------------------------------- Equation - 1
  2. CO32- (aq) + H2O (l)   \rightleftharpoons HCO3- (aq)   + OH- (aq) ----------- Equation - 2

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

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

​​​​​​​ CO32- (aq) + H2O (l)   \rightleftharpoons HCO3- (aq)   + OH- (aq)

\RightarrowPart - (b):

  • Step - 1:

We are given with the following:

  1. Ka,1 =   4.3 x 10-7
  2. Ka,2 = 5.6 x 10-11

​​​​​​​We should consider the value of Ka,2 , as the value of acid dissociation constant for H2CO3 ... at the above conditions...

  • Step - 2:

​​​​​​​We know,for H2CO3, --- Ka,2 = 5.6 x 10-11 , and ---  Kw = ( Ka,2) x ( Kb ) = 1.0 x 10-14

And we know from Equation - 2:

Kb = [ OH- ] x [ HCO3- ] / [ CO3 2- ] = Kw / Ka,2 = ( 1.0 x 10-14 ) / ( 5.6 x 10-11 ) = 1.786 x 10-4( approx.)

Therefore:

  • Answer(b): The value of Kb = 1.786 x 10-4( approx.)

​​​​​​​\RightarrowPart - (C):

  • Step - 1:

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

  CO32- (aq) + H2O (l)   \rightleftharpoons HCO3- (aq)   + OH- (aq) ----------- Equation - 2

  • Step - 2:

​​​​​​​We know the initial concentration of CO32- : [ CO32-]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:

Kb = [OH- ] x [HCO3-] /  [ CO32-] = 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)

\RightarrowPart - (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 )

.

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