Q: Concentration of H 2 SO 4 and Na 2 SO 4 in a solution is 1 M and 1 . 8 × 10 - 2 M , respectively. Molar solubility of PbSO 4 in the same solution is X × 10 - Y M (expressed in sc

H 2 SO 4 → H + + HSO 4 2 - 1 M 1 M ( K a 1 is very large) HSO 4 - ⇌ H + + SO 4 2 - K a 2 = 1 . 2 × 10 - 2 SO 4 2 - coming from Na 2 SO 4 = 1 . 8 × 10 - 2 SO 4 2 - H + HSO 4 - = 1 . 8 × 10 - 2 × 1 1 > K a 2 ∴ Rather than dissociation of HSO 4 - into H + and SO 4 2 - ions, association between already present H + and SO 4 2 - will take place. Assuming ' x ' mol / L of SO 4 2 - and H + combines to form HSO 4 - &#8

Q: A solution of 0 . 1 M weak base ( B ) is titrated with 0 . 1 M of a strong acid ( HA ) . The variation of pH of the solution with the volume of HA added is shown in the figure below. What is the pK b

At 3 ml pH of solution is 11 . B + HA → BH + + A − at 3 best buffer action is shown So, P OH = P K b = 3 (As basic buffer is formed) So, P K b = 3

Q: An acidified solution of 0 . 05 M Zn 2 + is saturated with 0 . 1 M H 2 S . What is the minimum molar concentration ( M ) of H + required to prevent the precipitation of ZnS

For ZnS K sp ( ZnS ) = Zn 2 + × S 2 - = 1 . 25 × 10 - 22 = 0 . 05 × S 2 - = 1 . 25 × 10 - 22 S 2 - = 25 × 10 - 22 For H 2 S H 2 S 0 .1 M ⇌ 2 H + + S 2 − 25 × 10 − 22 K a = H ' S 2 - H 2 S = 1 × 10 - 21 = H + 2 × 25 × 10 - 22 0 . 1 = 1 × 10 - 21 H + 2 = 1 25 H + = 1 5 H + = 0 . 2 M

Q: For the following reaction, the equilibrium constant K e at 298 K is 1.6 × 10 17 . F e 2 + a q + S 2 - a q ⇌ F e S s When equal volumes of 0.06 M F e 2 + a q and

Fe 2+ + S 2- ⇌ Fe S ( s ) ..... K c = 1.6 × 10 17 t = 0 : 0.03 M 0.1 M t = t; 0.03 - H 0.1 H Since K c > > 1 , n ≃ 0.03 ∴ [ S 2- ] = 0.1 - 0.031 = 0.07 M K c = Fe 2 + S 2- = 1 1.6 × 10 17 Fe 2 + = 1 1.6 × 0.07 × 10 - 17 y = 1 1.6 × 0.07 ≃ 8.93

Question 1

At 25^ C , the concentration of H ^ + ions in 1.00 10^ -3 M aqueous solution of a weak monobasic acid having acid dissociation constant (K_a )=4.00 10^ -11 is X 10^ -7 M . The value of X is _____ . Us

Accepted Answer

Because concentration of H + from weak acid is less we need to consider self ionization of H _2 O also. array rl HX ( aq ) & H ^ + + X ^ - ( aq ) \ 10^ -3 - x & x + y x \ H _2 O (l) & H ^ + ( aq )+ OH ^ - ( aq ) \ & x + y y array Approximation : (10^ -3 - x ) 10^ -3 aligned & x(x+y) 10^ -3 =K_a=4 10^ -11 ....(1) \ & y(x+y)=K w=10^ -14 .....(2) aligned array ll Add (1)+(2) \ array array ll & ( x + y )^2=5 10^ -14 \ & x + y = [ H ^ + ]= 5 10^ -7 \ & x = 5 =2.236 array Answer 2.23 or 2.24.

Question 2

The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL of 0.010 M barium nitrate with 100 mL of 0.10 M sodium iodate is X 10^ -6 ~mol dm ^ -3 . The value of X is . Use: Solub

Accepted Answer

[ NaIO _3 ]= 6 300 =2 10^ -2 M aligned Ba ( IO _3 )_2( ~s ) Ba ^ 2+ + & 2 IO _3^ - \ s & (2 10^ -2 +2 ~s ) aligned aligned & K _ sp = [ Ba ^ +2 ] [ IO _3^ - ]^2 \ & 1.58 10^ -9 = [ Ba ^ +2 ] (2 10^ -2 )^2 \ & [ Ba ^ +2 ]= s =3.95 10^ -6 M \ & X =3.95 aligned

Question 3

On decreasing the pH from 7  to 2 , the solubility of a sparingly soluble salt ( MX ) of a weak acid ( HX ) increased from 10 - 4 &#160; mol &#160; L - 1 to 10 - 3 &#160; mol &#160; L - 1 . The p

Accepted Answer

At  pH = 7 , i.e., the solubility in water is S , The solubility product,  K sp = S 2 &#160; &#160; - - - - - &#62; 1 Assume the solubility of sparingly soluble salt is  at &#160; pH = 2 MX &#8652; M &#8853; + X &#8854; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; S 1 &#160; &#160; &#160; &#160; &#160; &#160; &#160; S 1 - x The weak dissociation reaction is as follows, X &#8854; + H &#8853; &#8652; HX S 1 - x &#160; &#160; 10 - 2 &#160; &#160

Question 4

Concentration of H 2 SO 4  and Na 2 SO 4 in a solution is 1 M and 1 . 8 &#215; 10 - 2 M , respectively. Molar solubility of PbSO 4  in the same solution is X &#215; 10 - Y M (expressed in sc

Accepted Answer

H 2 SO 4 &#8594; H + + HSO 4 2 - 1 M 1 M  ( K a 1  is very large) HSO 4 - &#8652; H + + SO 4 2 - &#160;&#160; K a 2 = 1 . 2 &#215; 10 - 2 SO 4 2 - coming from Na 2 SO 4 = 1 . 8 &#215; 10 - 2 SO 4 2 - H + HSO 4 - = 1 . 8 &#215; 10 - 2 &#215; 1 1 &#62; K a 2 &#8756; Rather than dissociation of HSO 4 -  into H + and SO 4 2 - ions, association between already present H +  and SO 4 2 - will take place. Assuming ' x ' mol / L of SO 4 2 - and H + combines to form HSO 4 - &#8

Question 5

A solution is prepared by mixing 0 . 01 mol  each of H 2 CO 3 , NaHCO 3 , Na 2 CO 3 , and NaOH in 100 ~mL of water. pH of the resulting solution is [Given: pK a 1 and pK a 2 of H 2 CO 3 are 6 . 3

Accepted Answer

First acid base reaction between H 2 CO 3 and NaOH takes place. H 2 CO 3 0 . 01 mole + NaOH 0 . 01 mole &#10230; NaHCO 3 - 0 . 01 mole + H 2 O After the acid base reaction, we have 0 . 01 mole &#160; Na 2 CO 3   and 0 . 02  moles of NaHCO 3 . Here, This will form an acidic buffer of NaHCO 3 and Na 2 CO 3 . &#8756; &#160; pH = pK a 2 + log Salt Acid = 10 . 32 + log 0 . 01 0 . 1 0 . 02 0 . 1 = 10 . 32 + log 1 2 = 10 . 32 - log 2 = 10 . 32 - 0 . 3 = 10 . 02 &#8756; &#160; pH = 10 . 0

Question 6

A solution of 0 . 1 M weak base ( B ) is titrated with 0 . 1 M of a strong acid ( HA ) . The variation of pH of the solution with the volume of HA added is shown in the figure below. What is the pK b

Accepted Answer

At 3 &#160; ml &#160; pH of solution is 11 . B + HA &#8594; BH + + A &#8722; at 3 best buffer action is shown So,  P OH = P K b = 3  (As basic buffer is formed) So,  P K b = 3

Question 7

An acidified solution of 0 . 05 &#160; M &#160; Zn 2 + is saturated with 0 . 1 &#160; M &#160; H 2 S . What is the minimum molar concentration ( M ) of H + required to prevent the precipitation of ZnS

Accepted Answer

For  ZnS K sp ( ZnS ) = Zn 2 + &#215; S 2 - = 1 . 25 &#215; 10 - 22 = 0 . 05 &#215; S 2 - = 1 . 25 &#215; 10 - 22 S 2 - = 25 &#215; 10 - 22 For  H 2 S H 2 S 0 .1 &#8196; M &#8652; 2 H + + S 2 &#8722; 25 &#215; 10 &#8722; 22 K a = H ' S 2 - H 2 S = 1 &#215; 10 - 21 = H + 2 &#215; 25 &#215; 10 - 22 0 . 1 = 1 &#215; 10 - 21 H + 2 = 1 25 H + = 1 5 H + = 0 . 2 &#160; M

Question 8

For the following reaction, the equilibrium constant K e at 298 &#160; K is 1.6 &#215; 10 17 . F e 2 + a q + S 2 - a q &#160; &#8652; F e S s When equal volumes of 0.06 &#160; M &#160; F e 2 + a q and

Accepted Answer

Fe 2+ + S 2- &#8652; Fe&#8201; S ( s ) ..... K c = 1.6 &#215; 10 17 t = 0 : 0.03 &#8201;M&#160; 0.1 &#8201;M t = t; &#160; 0.03 - H &#160; 0.1 &#8201;H Since K c &#62; &#62; 1 , n &#8771; 0.03 &#8756; &#8201; [ S 2- ] = 0.1 - 0.031 = 0.07 &#8201;M K c = Fe 2 + S 2- = 1 1.6 &#215; 10 17 Fe 2 + = 1 1.6 &#215; 0.07 &#215; 10 - 17 y = 1 1.6 &#215; 0.07 &#8771; 8.93

Ionic Equilibrium — Previous Year Questions & Solutions