Solving Weak Acid and Weak Base pH problems | Chemistry Net

Solving Weak Acid and Weak Base pH problems

Solving Weak Acid and Weak Base pH problems


Solving Weak Acid and Weak Base pH problems




Fig. I.1: Flowchart showing how to calculate the pH of a weak acid solution
Fig. I.1: Calculating the pH of a weak acid
In a previous post entitled “Weak Acids and Bases – Calculate the pH of a weak acid” a general equation was derived for the calculation of [H+] in weak acid (base) solutions (mathematical approach).  A four-step method was also proposed for solving weak acid base problems and for the calculation of the pH of a weak acid or base (chemical approach) (Fig. I.1).
Some extra solved examples on weak acid (base) chemistry are shown below. The “chemical approach” method is used for the solution.

 

Example I.1


What the concentration of an acidic solution ΗΑ must be so that 
 [H+]= 3.5 * 10-4 M (ka =1,7 * 10-5)

STEP
RESULT




GIVEN



[H+]= 3.5 * 10-4 M
ka =1.7 * 10-5
UNKNOWN
[HΑ] = ;

I

HA is a weak acid as the ka value shows.
Write the ionization equilibrium reaction:
   
          HΑ         Η+    +   Α-            (1)   

                                                    

II

Write down the equilibrium constant expression and the equilibrium constant:

ka =  +] . [A-] / [HA] = 1.7 * 10-5   (2)


III

Let us suppose that y M is the initial concentration of HA. Then the equilibrium concentrations of the species involved are as follows:

Initial
y Μ
0 Μ
0 Μ
Change
-3.5 * 10-4 
+3.5 * 10-4 
+3.5 * 10-4 
Final
(at equilibrium)
(y – 3.5 * 10-4)
3.5 * 10-4 
3.5 * 10-4 

 IV

From (2) and the equilibrium concentrations from the above table:
ka = [Η+] . [A-] / [HA] = (3.5 * 10-4)2  / (y-3.5 * 10-4)  = 1.7 * 10-5   
and y = 7.2 * 10-3 M   
Therefore  [HΑ] = 7.2 * 10-3 M  



 

Example I.2


A weak acid solution is prepared by dissolving 8.3 * 10-1 g of HA in water. The final volume of the solution is measured to be 100 ml. What is the equilibrium concentration of the species involved? (kΗΑ = 1.8 * 10-4, ΜΒΗΑ = 46 g/mol).

STEP
RESULTS



DATA


 mHA = 8.3 * 10-1 g
VΔ = 100 ml
kHA = 1.8 * 10-4
ΜΒΗΑ = 46 g/mol
UNKNOWN
[HΑ] = ?  +] = ? 
[A-] = ?



I

HA is a weak acid as we can see from kHA. Therefore, it dissociates partially in water.
Write the ionization equilibrium reaction:
   
          HΑ         Η+    +   Α-            (1)   

Calculate the moles of  ΗΑ in solution: 
nHA = 8.3 * 10-1 g /46 g/mol  = 1.8 * 10-2 mol

The initial concentration of ΗΑ is calculated as follows: 
      100 ml solution  contain  1,8 . 10-2 mol ΗΑ
     1000 ml solution contain      y = ; mol ΗΑ 
y = [HΑ]initial =1.8 * 10-1 mol/ℓ   (2)



II

Write down the equilibrium constant expression and the equilibrium constant:

kHA =  +] . [A-] / [HA] = 1,8 . 10-4   (3)



III

Let us suppose that x moles ΗΑ dissociate. The equilibrium concentration of the species involved is given below:
                                           

HΑ
Η+
Α-          
Initial
0.18 Μ
0
0
Change
-x M
+x M
+x M
Final (at equilibrium)
(στην χημ. ισορρ.)
(0.18-x) M
+x M
+x M


IV

From (3) and the equilibrium concentrations of the species involved:
ka =  [Η+] . [A-] / [HA] = (x)2 / (0,18-x ) = 1.8 * 10-4  
  (x)2 /0.18 = 1.8 * 10-4 
x = 5.7 * 10-3 M  (4)  (assuming that 0,18-x ≈ 0,18 since ΗΑ is a weak acid and it partially dissociates in water. Therefore the value of x will be small compared to the initial concentration of ΗΑ)

Checking for the validity of the above approximation: If the value of x is less than 5% of the initial concentration of HA the approximation 0,18-x ≈ 0,18 is valid.

5.7 * 10-3 * 100 / 0.18  = 3.2%





Example I.3


The ionization constant of NH3 in water is 10-5. Calculate the following for a 0.1     M NH3 solution: a) the degree of dissociation α  b) the concentration of [OH-] in solution

STEP
RESULTS



DATA

 ΝΗ3
kb = 10-5
[ΝΗ3] = 0.1 M
UNKNOWNS
α = ?  [ΟΗ-] = ?


I
ΝΗ3 is a weak base as the value of kb shows. Therefore, it dissociates partially in water according to:

H2O   +      ΝΗ3           ΝΗ4+    +    ΟΗ-   (1)

Let us suppose that α is the degree of dissociation of ΝΗ3 in water. Therefore:
α = moles ΝΗ3 that dissociate / total moles ΝΗ3
Therefore moles ΝΗ3 that dissociate = 0.1 * α    (2)


II
Write down the equilibrium constant expression and the equilibrium constant:

kb =  [NΗ4+] . [OH-] / [NH3] = 10-5   (3)


III

The equilibrium concentration of the species involved is given below:


ΝΗ3
ΝΗ4+
ΟΗ-   
Initial
0.1 Μ
0
0
Change
(0.1-0.1α)  M
+0.1 * α M
+0.1 * α M
Final
(at equilibrium)
0.1 * (1-α) M
0.1 * α M
+0.1 * α M


IV
From (3) and the above equilibrium concentrations of the species involved:
Kb = (0.1 * α). (0.1 * α) / 0.1 * (1-α) = 10-5 = 0.01 * α2 / 0.1 = 10-5  
Therefore,  α = 0,01 (4) 
Assuming that (1-α) ≈ 1 since the value of α is small compared to [ΝΗ3] = 0.1 M   

Therefore [OH-] = 0.1 *α = 0.1 * 0.01 = 10-3 M

Therefore  α = 0.01 and [OH-] = 10-3 M


 

Relevant Posts




References

J-L. Burgot “Ionic Equilibria in Analytical Chemistry”, Springer Science & Business Media, 2012
J.N. Butler  “Ionic Equilibrium – Solubility and pH calculations”, Wiley – Interscience, 1998
Clayden, Greeves, Waren and Wothers “Organic Chemistry”, Oxford,
D. Harvey,  “Modern Analytical Chemistry”, McGraw-Hill Companies Inc., 2000
Toratane Munegumi, World J. of Chem. Education, 1.1, 12-16 (2013)
J.N. Spencer et al., “Chemistry structure and dynamics”, 5th Edition, John Wiley & Sons, Inc., 2012
L. Cardellini, Chem. Educ. Res. Pract. Eur., 1, 151-160, (2000)




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