Question 1

In a qualitative analysis Bi ^ 3+ is detected by appearance of precipitate of BiO(OH)(s) . Calculate pH when the following equilibrium exists at 298 K : BiO(OH)(s) BiO ^ + (aq) + OH ^ - (aq) , K = 4 1

Accepted Answer

The equilibrium reaction is given by: BiO(OH)(s) BiO ^ + (aq) + OH ^ - (aq) The equilibrium constant expression is: K = [ BiO ^ + ][ OH ^ - ] Let the solubility of BiO(OH) be s. Then [ BiO ^ + ] = s and [ OH ^ - ] = s. Substituting the values into the equilibrium expression: s^2 = 4 10^ -10 s = 2 10^ -5 M Therefore, the concentration of hydroxide ions is: [ OH ^ - ] = 2 10^ -5 M Calculating the pOH: pOH = - [ OH ^ - ] = - (2 10^ -5 ) = 5 - 2 Given 2 = 0.3010: pOH = 5 - 0.3010 = 4.699 The pH of t

Question 2

At 298 K, a certain buffer solution contains equal concentrations of X^ - and HX, K_b for X^ - is 10^ -10 . What is the pH of this buffer solution ?

Accepted Answer

Given K_b for X^ - = 10^ -10 . For a conjugate acid-base pair at 298 K, K_a K_b = K_w = 10^ -14 . K_a = 10^ -14 10^ -10 = 10^ -4 The pK_a of the weak acid HX is: pK_a = - (K_a) = - (10^ -4 ) = 4 Using the Henderson-Hasselbalch equation for an acidic buffer: pH = pK_a + ( [X^ - ] [HX] ) Since the concentrations of X^ - and HX are equal, [X^ - ] = [HX]. pH = 4 + (1) = 4 + 0 = 4 Answer: 4

Question 3

The correct order of solubility of the given salts in water at 298 K is Salt K_ sp at 298 K AgBr 5.0 10^ -13 Zn(OH)_2 1.0 10^ -15 Hg_2Cl_2 1.3 10^ -18

Accepted Answer

Let s be the solubility of the salt in water. For AgBr : AgBr Ag ^+ + Br ^- K_ sp = s^2 s = 5.0 10^ -13 = 50 10^ -14 7.07 10^ -7 M For Zn(OH) _2: Zn(OH) _2 Zn ^ 2+ + 2 OH ^- K_ sp = s(2s)^2 = 4s^3 s = ( 1.0 10^ -15 4 )^ 1/3 = (250 10^ -18 )^ 1/3 6.3 10^ -6 M For Hg _2 Cl _2: Hg _2 Cl _2 Hg _2^ 2+ + 2 Cl ^- K_ sp = s(2s)^2 = 4s^3 s = ( 1.3 10^ -18 4 )^ 1/3 = (0.325 10^ -18 )^ 1/3 6.87 10^ -7 M Comparing the solubilities: 6.3 10^ -6 > 7.07 10^ -7 > 6.87 10^ -7 Thus, the correct order of solubility

Question 4

Phenolphthalein is used as an indicator for the titration of sodium hydroxide solution against a standard solution of oxalic acid. The colour change that is observed at an alkaline pH close to the equ

Accepted Answer

In the standardisation of sodium hydroxide using a standard solution of oxalic acid, oxalic acid (a primary standard) is taken in the conical flask and sodium hydroxide is taken in the burette. Phenolphthalein is used as the indicator, which is colourless in an acidic medium and pink in an alkaline medium. Initially, the solution in the conical flask contains oxalic acid, so it remains colourless. As sodium hydroxide is added, the pH of the solution increases. At the equivalence point, the salt

Question 5

Phosphoric acid ionizes in three steps with their ionization constant values K _ a _1 , ~K _ a _2 and K _ a _3 , respectively, While K is the overall ionization constant. Which of the following statem

Accepted Answer

For polybasic acid, order of ionization constant is Ka _1 Ka _2 Ka _3 because as negative charge increases, tendency to loose H ^ + ions decreases. Overall K = Ka _1 Ka _2 Ka _3 Taking log on both sides K = Ka _1+ Ka _2+ Ka _3

Question 6

The ratio of solubility of AgCl in 0.1 M KCl solution to the solubility of AgCl in water is: (Given : Solubility product of AgCl =10^ -10 )

Accepted Answer

Answer option index: 0

Question 7

Which indicator is used in the titration of sodium hydroxide against oxalic acid and what is the colour change at the end point?

Accepted Answer

For weak acid and strong base titration, phenolphthalein is used as indicator. In acidic medium phenolphthalein is colourless while in alkaline medium it is pink in colour.

Ionic Equilibrium — Previous Year Questions & Solutions