Chemistry
Download
1 / 56

Chemistry - PowerPoint PPT Presentation


  • 119 Views
  • Uploaded on
  • Presentation posted in: General

Chemistry. Ionic equilibrium-II. Session Objectives. Session Objectives. pH of weak acids p H of mixture of two strong acids p H of mixture of strong and weak acids D issociation of polybasic acids p H of mixture of two weak acids. For mixture of two strong acids.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha

Download Presentation

Chemistry

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript





Session objectives
Session Objectives

  • pH of weak acids

  • pH of mixture of two strong acids

  • pH of mixture of strong and weak acids

  • Dissociation of polybasic acids

  • pH of mixture of two weak acids


For mixture of two strong acids
For mixture of two strong acids

Let us consider a mixture containing 200 ml 0.1 M HCl and 500 ml 0.2 M H2SO4.Since both are strong electrolytes,

200 × 0.1×10–3 0.02 mole

= 0.02 mole

500 × 0.2 × 10–3 2 × 0.1 mole

= 0.1 mole


For mixture of two strong acids1

Total ion concentration,

For mixture of two strong acids

Note: For mixture of any two acids or bases, degree of ionization of water is taken negligible because of common ion effect.


Mixture of strong and weak acids
Mixture of strong and weak acids

Let us consider a mixture of strong acid(HA) and a weak acid (HX) of concentration C1 and C2 respectively.

Now, for strong acid

For weak acid,

Dissociation constant of weak acid,




Illustrative example 1

H2S H+ + HS–

Illustrative example 1

A solution contains 0.10 M H2S and0.3 M HCl. Calculate the concentration of [HS–] and [S–2] ions in the solution.For H2S,Ka1 = 1.0 x 10-7Ka2 = 1.3 x 10-13

Solution:


Solution

HS– H+ + S–2

Solution


Solution1
Solution

Considering [HS–] dissociates to a very small extent


Dissociation of polybasic acids
Dissociation of polybasic acids

Acids giving more than one hydrogen ion per molecule are ‘polybasic’ or ‘polyprotic’ acids.

Examples are H2C2O4, H2CO3, H2S, H3PO4, H3AsO4, etc.

These dissociate in stages . For example,


Dissociation of polybasic acids1

Normally K2 << K1.To calculate hydrogen ion concentration, only the first step should be consideredas the H+ obtained from successive dissociation can be neglected, but to calculate the concentration of

then both the equilibria have to be considered.

Dissociation of polybasic acids


Ph of mixture of two weak acids
pH of mixture of two weak acids

Let HA and HB are two weak acids.


Ph of mixture of two weak acids1
pH of mixture of two weak acids

Dividing (1) by (2)



Illustrative example 2
Illustrative example 2

A solution is prepared by mixing 0.2MHCOOH with 0.5 M CH3 COOH.

Given KaCH3COOH=1.8 x 10–5 , KaHCOOH =2.1x10-4

Calculate [HCOO–] , [CH3COO–] and pH of the solution.

Solution:


Solution2
Solution

From (2)




Hydrolysis of salts
Hydrolysis of Salts

A. Hydrolysis of a salt of weak acid and strong base

The hydrolysis reaction is

At eqm.

where C = concentration of salt

h = degree of hydrolysis.


Hydrolysis of salts1
Hydrolysis of Salts

Hydrolysis constant



Hydrolysis of salts3
Hydrolysis of Salts

B. Salt hydrolysis of strong acid and weak base

where Kb = Dissociation constant of weak base.


Hydrolysis of salts4
Hydrolysis of Salts

C. Hydrolysis for a salt of weak acid and weak base

At eqm.

At eqm.



Hydrolysis of salts6
Hydrolysis of Salts

Now, to calculate the pH

For pH,



Illustrative example 3
Illustrative example 3

Calculate the pH at the equivalence point of the titration between 0.1 M CH3COOH (50 ml) with 0.05 M NaOH . Ka(CH3COOH) = 1.8 × 10–5.

Solution:

At the equivalence point,

Let V ml NaOH is required to reach the equivalence point.

At the equivalence point,


Solution5
Solution

V = 100 ml



Illustrative example 4
Illustrative example 4

Calculate the percentage hydrolysis of decinormal solution of ammonium acetate, given Ka = 1.75 × 10–5, Kb = 1.80 × 10–5and Kw = 1 × 10–14. What will be the change in degree of hydrolysis when 2 L of water is added to 1 L of the above solution?

Solution:

Since CH3COONH4 is a salt of weak acid and weak base


Solution7
Solution

Since degree of hydrolysis is not affected by the concentration in this case. So on dilution there will be no change in degree of hydrolysis.


Illustrative example 5
Illustrative example 5

Hydrolysis constant of Zn+2 is 1 × 10–9

(a) Calculate pH of a 0.001 M solution of ZnCl2.

(b) What is the basic dissociation constant of Zn(OH)+?

Solution:



Illustrative example 6
Illustrative example 6

When 0.20 M acetic acid is neutralised with 0.20 M NaOH in 0.50 litre of water, the resulting solution is slightly alkaline, calculate the pH of the resulting solution (Ka for acetic acid = 1.8 x 10-5)

Solution:



Illustrative example 7
Illustrative example 7

Calcium lactate is a salt of a weak organic acid and represented as Ca(Lac)2. A saturated solution of Ca (Lac)2 contains 0.13 mol of this salt in 0.50 litre solution. The pOH of this solution is 5.60. Assuming a complete dissociation of the salt, calculate Ka of lactic acid.

Solution:




Class exercise 1

The hydrolysis constant for FeCl2 will be

Class exercise 1

Solution:

FeCl2 is the salt of strong acid and weak base.


Solution11
Solution

Hence, the answer is (b)


Class exercise 2
Class exercise 2

pH of a solution produced when an aqueous solution of pH 6 is mixed with an equal volume of an aqueous solution of pH 3, will be

(a) 4.5(b) 4.3(c) 4.0(d) 3.3

Solution:

pH = 6

pH = 3


Solution12
Solution

= 4 – log 5.005

= 3.3

Hence, the answer is (d).


Class exercise 3

Which one of the following is true forany diprotic acid, H2X?

(a) Ka2 > Ka1(b) Ka1 > Ka2(c) Ka1 = Ka2

Class exercise 3

Solution:

H2X being a diprotic acid,

Due to the ‘common ion effect’ dissociation of HX–will be less.


Class exercise 4
Class exercise 4

Ka (CH3COOH) = 1.7 × 10–5 and [H+] = 3.4 × 10–4. Then initial concentrations of CH3COOH is

(a) 3.4 × 10–4(b) 6.8 × 10–3(c) 3.4 × 10–3(d) 6.8 × 10–2

Solution:


Solution13
Solution

Hence, the answer is (b).


Class exercise 5
Class exercise 5

0.001 M HCl is mixed with 0.01 M HCOOH at 25° C. If Ka HCOOH = 1.7 × 10–4, find the pH of the resulting solution.

Solution:



Class exercise 6

What is the percentage hydrolysis of NaCN solution when the Ka HCN = 1.3 × 10–9,K2 = 1 × 10–14?

(a) 2.48(b) 5.26(c) 9.6(d) 8.2

Class exercise 6

Solution:


Solution15
Solution

= 2.48 × 10–2


Class exercise 7
Class exercise 7

Calculate the pH of a solution obtained by mixing 100 ml of 0.1 M HCl with 9.9 ml of 1 M NaOH.

Solution:

Acid remaining = 1 × 10–4 moles

pH = –log[H+] = 4 – log 9.09

= 3.04


Class exercise 8
Class exercise 8

Calculate the percentage hydrolysis of 3 × 10–3 M aqueous solution of NaOCN (Ka HCON = 3.33 × 10–4 M).

Solution:




ad
  • Login