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Showing posts with label IONIC EQUILIBRIUM:. Show all posts
Showing posts with label IONIC EQUILIBRIUM:. Show all posts

Saturday, September 26, 2020

The pH of at which an acid indicator with Ka is 10^-15 changes colour when indicator concentration 1×10^-5 M is?


SOLUTION:

An indicator with Ka = 10^-5 is solution with pH = 6 Calculate % of indicator in ionised form?


SOLUTION:

For acidic indicator, pH range is 3 to 4.6 calculate the ratio of[ In- ] and H+ for the appearance of solution in a single colour.


SOLUTION:
Given pH range 3.0 to 4.6 so pKIn= (3.0+ 4.6)/2 =3.6

For an indicator pKa is 6 Calculate pH of Solution having this indicator such that 40% indicator molecules remain in ionised form.

SOLUTION:  We know that  

The pH range of a basic indicator is 4 to 6.5 Calculate the dissociation constant of indicator?.

SOLUTION:  pKIn must be midpoint of pH range for acidic indicators and pOH range for basic indicators
The pH range = 4 to 6.5 so pOH range is 10 to 7.5
Hence PkIn =   (10+7.5) /2 = 8.75 

What are DOUBLE SALTS ?

DOUBLE SALTS:
(1)The addition compounds formed by the combination of two simple salts are termed as double salts.
(2) Double salts are stable in solid state only.
(3) When dissolved in water, it furnishes all the ions present in the simple salt form which it has been constituted.
(4)The solution of double salt shows the properties of the samples salts from which it has been constituted
For examples
Mohar’s salt-FeSO4 (NH4)2SO4 .6H2O (Ferrous ammonium sulphate)
Alum’s- K2SO4Al2 (SO4)3. 24H2O (Potassium ammonium sulphate)
Karnalite- KCl.MgCl2.6 (H2O)
Dolomite- CaCO3.MgCO3 or CaMg (CO3)2

What is the pH of 0.10 M CH3COONa solution. Hydrolysis constant of sodium acetate is 5.6 × 10-10 ?

SOLUTION: Hydrolysis of the salt may be represented as

A 0.0258 M solution of the sodium salt, NaH of the weak monoprotic acid, HA has a pH of 9.65. Calculate Ka of the acid AH.


Friday, November 29, 2019

TITRATION OF WEAK ACID WITH STRONG BASE:


TITRATION OF ACETIC ACID VS SODIUM HYDROXIDE: 
ILLUSTRATIVE EXAMPLE: Give the answers of following questions when 20 ml of acetic acid (CH3COOH) is titrated with 0.10 M NaOH the (Given that Ka=2×10-5).
(A) Write out the reactions and equilibrium expression associated with Ka.
(B) Calculate the PH when:
(1) 20 ml of 0.10M CH3COOH + 0.0 ml of 0.10M NaOH 
(2) 20 ml of 0.10M CH3COOH + 5.0 ml of 0.10M NaOH
(3) 20 ml of 0.10M CH3COOH + 10 ml of 0.10M NaOH
(4) 20 ml of 0.10M CH3COOH + 15 ml of 0.10M NaOH
(5) 20 ml of 0.10M CH3COOH + 19 ml of 0.10M NaOH
(6) 20 ml of 0.10M CH3COOH + 20 ml of 0.10M NaOH
(7) 20 ml of 0.10M CH3COOH + 21 ml of 0.10M NaOH
(8) 20 ml of 0.10M CH3COOH + 25 ml of 0.10M NaOH
(9) 20 ml of 0.10M CH3COOH + 20 ml of 0.10M NaOH

SOLUTION:

(A) Write out the reactions and equilibrium expression associated with Ka.
(B) Calculate the PH when:

S.N.
Given condition
comments
PH
1
20 ml of 0.10M CH3COOH + 0.0 ml of 0.10 ml NaOH 
WA, PH=1/2(Pka-logC)
2.85
2
20 ml of 0.10M CH3COOH + 5.0 ml of 0.10 ml NaOH 
ABS, PH=PKa+ log[S]\[A]
4.22
3
20 ml of 0.10M CH3COOH + 10 ml of 0.10 ml NaOH 
BB , PH=PKa
Half of equivalent point
4.70
4
20 ml of 0.10M CH3COOH + 15 ml of 0.10 ml NaOH 
ABS, PH=PKa+ log[S]\[A]
5.17
5
20 ml of 0.10M CH3COOH + 19 ml of 0.10 ml NaOH 
ABS, PH=PKa+ log[S]\[A]
5.98
6
20 ml of 0.10M CH3COOH + 20 ml of 0.10 ml NaOH 
SH, PH=7+ 1/2(Pka-logC)
Equivalent point
8.7
7
20 ml of 0.10M CH3COOH + 21 ml of 0.10 ml NaOH 
Strong Base
11.39
8
20 ml of 0.10M CH3COOH + 25 ml of 0.10 ml NaOH 
Strong Base
12.04
9
20 ml of 0.10M CH3COOH + 30 ml of 0.10 ml NaOH 
Strong Base
12.30


(C) Sketch the titration curve for this titration.

Wednesday, November 27, 2019

TITRATION OF STRONG ACID WITH STRONG BASE:


The titration of HCl (aq) with a standardized NaOH solution illustrated the titration of strong acid by a strong base.
The molecular and net ionic equation is.
Case (1): At the start point before any titrant has been added the receiving flask contains only 0.10 M HCl and 50 ml. Because it is strong acid so
Case (2): After starting but before equivalent point.
Case (3): At equivalent point
Case (4): Before equivalent point
TITRATION SUMMARY TABLE:

S.N.
Volume of
HCl Taken
Volume
of NaOH
PH

1
50.0 ml (In ml)
And 0.10 M
0.0 (In ml)
And 0.10M
1.0

2

10
1.17

3

20
1.36

4

30
1.60
NAVA>NBVB
5

40
1.95

6
(Vertical Over)
45
2.27

7

49
2.99

8
50.0 ml (In ml)
And 0.10 M
50
7.0
NAVA=NBVB
9

51
11
NAVA<NBVB
10

60
11.95























GRAPHICAL REPRESENTATION:
ILLUSTRATIVE EXAMPLE: Find the pH of following titrations:
(A) 500 ml, 0.10 M HCl + 500 ml 0.10 M Ca(OH)2
(B) 400 ml, M/200 Ca(OH)2 + 400 ml M/50 HNO
 ANSWERS KEY:
(A): PH=12.6989  (B): PH=2.6