What will be the pH of the buffer solution containing 0.15 moles of NH4OH and 0.25 moles of NH4Cl Kb for NH4OH is 1.8x10^-5.

 

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 pK_In= (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:  pK_In 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 Pk_In =   (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.MgCl_2.6 (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.

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.

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 HNO₃ 

 _{ANSWERS KEY:
(A): PH=12.6989  (B): PH=2.6}