Titration of Borax: Tincal : Suhaga :

When Borax dissolved in water, it is subject to completely dissociate, and then, hydrolyzes to orthoboric acid (B(OH)₃  and [B(OH)₄]^- anions,  and resulting PH  of  solution about 9.13.

According to the equation:

Na₂[B₄O₅(OH)₄]⋅8H₂O(s)⟶2Na⁺(aq) + B₄O₅(OH)₄²⁻ (aq) + 8H₂O(l)

B₄O₅(OH)₄²⁻(aq) + 5H₂O(l) ⟶ 4 B(OH)₃(aq) + 2OH⁻(aq)

The liberated B(OH)₃(aq)  acts as a Lewis acid on OH⁻ ions from auto ionization of water, and hence, in equilibrium with water:

B(OH)₃(aq)+2H₂O(l) ⟶ B(OH)₄⁻(aq) + H₃O⁺(aq)

Therefore, one mole of borax reacts with two moles of strong acid solutions. This is  because when Borax dissolved in water Both B(OH)₃(aq)  and B(OH)₄⁻(aq) are formed but only  B(OH)₄⁻(aq) react with acid.  

For example:

                Na₂[B₄O₅(OH)₄].8H₂O(aq) + 2HCl(aq) ⟶ 4B(OH)₃(aq)+ 2NaCl (aq) + 5 H₂O

ILLUSTRATIVE EXAMPLE: Aqueous solution of borax reacts with two moles of acids.

This is because of:

(A) Formation of 2mol of B(OH)₃ only

(B) Formation of 2mol of Na[B(OH)₄]¯ only

(C) Formation of 1mol each of B(OH)₃ and Na[B(OH)₄]¯

(D) Formation of 2mol each of Na[B(OH)4]¯and B(OH)₃ of which only Na[B(OH)₄]¯ reacts with acid.

Answer: (D)

Solution: 

 Na₂B₄O₇ + 7H₂O ⟶ 2B(OH)₃ + 2Na[B(OH)₄]

B(OH)₃ or H₃BO₃ is an acid and does not react with acid. Hence Na[B(OH)₄] reacts with acid.

What is structure of Borax: Tincal or Suhaga ?

The name borax is derived from the Persian word borak, meaning white. IUPAC name of borax is sodium tetraborate decahydrate (Na₂B₄O₇⋅10 H₂O ). Also note that, after analyzing its crystal structure, borax is more properly formulated as Na₂[B₄O₅(OH)₄]⋅8H₂O

Structure:

(1) It is also named as sodiumtetra borate decahydrate

(2) Common Indian name is Suhaga.

(3) Na₂B₄O₇. 5H₂O is known as Jeweller's Boran

(4) Na₂B₄O₇ is known as Boron glass

(5) It is also represent as Na₂O.2B₂O₃.10H₂O.

IIT UPDATE: (2016):

Titration of Boric Acid:

Boric acid is a weak monoprotic acid (Ka=6.4×10⁻¹⁰). Therefore, it cannot be directly titrated with standard strong alkali (NaOH). But by the addition of an organic poly hydroxy compound such as 1,2- or 1,3-diol, (Like Sugar, Mannitol, Catechol)  it is converted to a much stronger acid, which can be titrated using phenolphthalein. This is due to the complex formation between hydrated borate ion and either 1,2- or 1,3-diol  for example .

Ploy hydroxy (Cis-diol) increases acidity of boric acid due to formation of complex with metaborate and reaction goes to farward to produce more H⁺.

IIT UPDATE: (2006):

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}