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CHXI07:EQUILIBRIUM

314894 An aqueous solution contains an unknown concentration of ${Ba}^{2 +}$ . When $50 ml$ of a $1 M$ solution of ${Na}_{2} {SO}_{4}$ is added, ${BaSO}_{4}$ just begins to precipitate. The final volume is $500 mL$ . The solubility product of ${BaSO}_{4}$ is $1 \times 10^{- 10}$ . What is the original concentration of ${Ba}^{2 +}$ ?

1

2 \times 1 0^{- 9} M

2

1 .1 \times 1 0^{- 9} M

3

1 .0 \times 1 0^{- 10} M

4

5 \times 1 0^{- 9} M

Explanation:

$50 m L$ $1 M$ $N a_{2} S O_{4}$ is made to $500 mL$
means ${BaSO}_{4}$ solution is of $450 ml$ .
Concentration of ${SO}_{4}^{- 2}$ in ${Ba}^{+ 2}$ solution
$M_{1} V_{1} = M_{2} V_{2} 1 \times 50 = M_{2} \times 500 M_{2} = \frac{1}{10}$
for just precipitation
Ionic product $= K_{S P}$
$[B a^{+ 2}] [{SO}_{4}^{- 2}] = K_{SP} (BaS O_{4}) [B a^{+ 2}] \times \frac{1}{10} = 1 0^{- 10}$
$[B a^{+ 2}] = 1 0^{- 9} M in 500 ml solution$
for calculation of $[B a^{+ 2}]$ in original solution (450ml)
$M_{1} \times 450 = 10^{- 9} \times 500 M_{1} = \frac{500}{450} \times 10^{- 9} = 1.11 \times 10^{- 9} M$
$[M_{1} =$ molarity of ${Ba}^{+ 2}$ in original solution (450 $ml)$ ]

CHXI07:EQUILIBRIUM

314895 When equal volumes of the following solutions are mixed, precipitation of $AgCl (K_{sp} = 1.8 \times 10^{- 10})$ will only with

1

1 0^{- 4} M A g^{+} and 1 0^{- 4} M C l^{-}

2

1 0^{- 6} M A g^{+} and 1 0^{- 6} M C l^{-}

3

1 0^{- 7} M  A g^{+} and 1 0^{- 7} M C l^{-}

4

1 0^{- 5} M  A g^{+} and 1 0^{- 5} M C l^{-}

Explanation:

$K_{S P} = [{Ag}^{+}] [{Cl}^{-}]$
to get ppt Ionic product $> K_{S P}$
$1.8 \times 10^{- 10} = 10^{- 4} \times 10^{- 4} = 10^{- 8}$
$10^{- 8} > 1.8 \times 10^{- 10}$ .

CHXI07:EQUILIBRIUM

314896 The mass of ${Pb}^{2 +}$ ion is left in solution when $50 mL of 0 .2 mL of 0 .2 M Pb(N O_{3})_{2}$ is added to $50 mL$ of 1.5 M NaCl is (Given $K_{SP}$ for $PbC l_{2}$ = $1 .7 \times 1 0^{- 4} M^{3 -}$

1

12 g

2

12 mg

3

120 g

4

120 mg

Explanation:

Assume complete precipitation of the ${Pb}^{2 +}$
followed by solution of the equilibrium
concentration to be determined.
Number of millimoles of
${Pb}^{2 +} = (50 mL) (0.2 mmol / mL) = 10 mmol$
Number of millimoles of
${Cl}^{-} = (50 mL) (1.5 mmol / mL) = 75 mmol$
${Pb}^{2 +} + 2 {Cl}^{-} \to {PbCl}_{2}$
What is the lead concentration
that can exist in a solution of
$C l^{-}$ containing $75 mol - 20mmol$
$= 55 m mol$ in $100 mL$
Concentration of
$C l^{-} = 55 m mol 100 mL = 0 . 55 M$
${PbCl}_{2} ⇌ {Pb}^{2 +} + 2 {Cl}^{-}$
$K_{s p}$ = $P b^{2 +} {[C l^{-}]}^{2} = 1 .7 1 0^{- 4} M^{3}$
$[P b^{2 +}] = \frac{1.7 10^{- 4}}{(0.55)^{2}} = 5.6 \times 10^{- 4} M$
$\frac{w_{2} \times 1000}{M W_{2} \times V (m L)} = 5.6 \times 10^{- 4}$
$\frac{w_{2} \times 1000}{208 \times 100} = 5.6 \times 10^{- 4}$ weight of ${Pb}^{2 +}$ ion $= 12 mg$

CHXI07:EQUILIBRIUM

314897 In the third group of qualitative analysis, the precipitating reagent is ${NH}_{4} Cl + {NH}_{4} OH$ . The function of ${NH}_{4} Cl$ is to

1 Increase the ionization of

{NH}_{4} OH

2 Supress the ionization of

{NH}_{4} OH

3 Stabilise the hydroxides of group cations

4 Convert the ions of group III into their respective chlorides

Explanation:

Due to common ion effect, ${NH}_{4} OH$ ionization is suppressed.

CHXI07:EQUILIBRIUM

314894 An aqueous solution contains an unknown concentration of ${Ba}^{2 +}$ . When $50 ml$ of a $1 M$ solution of ${Na}_{2} {SO}_{4}$ is added, ${BaSO}_{4}$ just begins to precipitate. The final volume is $500 mL$ . The solubility product of ${BaSO}_{4}$ is $1 \times 10^{- 10}$ . What is the original concentration of ${Ba}^{2 +}$ ?

1

2 \times 1 0^{- 9} M

2

1 .1 \times 1 0^{- 9} M

3

1 .0 \times 1 0^{- 10} M

4

5 \times 1 0^{- 9} M

Explanation:

$50 m L$ $1 M$ $N a_{2} S O_{4}$ is made to $500 mL$
means ${BaSO}_{4}$ solution is of $450 ml$ .
Concentration of ${SO}_{4}^{- 2}$ in ${Ba}^{+ 2}$ solution
$M_{1} V_{1} = M_{2} V_{2} 1 \times 50 = M_{2} \times 500 M_{2} = \frac{1}{10}$
for just precipitation
Ionic product $= K_{S P}$
$[B a^{+ 2}] [{SO}_{4}^{- 2}] = K_{SP} (BaS O_{4}) [B a^{+ 2}] \times \frac{1}{10} = 1 0^{- 10}$
$[B a^{+ 2}] = 1 0^{- 9} M in 500 ml solution$
for calculation of $[B a^{+ 2}]$ in original solution (450ml)
$M_{1} \times 450 = 10^{- 9} \times 500 M_{1} = \frac{500}{450} \times 10^{- 9} = 1.11 \times 10^{- 9} M$
$[M_{1} =$ molarity of ${Ba}^{+ 2}$ in original solution (450 $ml)$ ]

CHXI07:EQUILIBRIUM

314895 When equal volumes of the following solutions are mixed, precipitation of $AgCl (K_{sp} = 1.8 \times 10^{- 10})$ will only with

1

1 0^{- 4} M A g^{+} and 1 0^{- 4} M C l^{-}

2

1 0^{- 6} M A g^{+} and 1 0^{- 6} M C l^{-}

3

1 0^{- 7} M  A g^{+} and 1 0^{- 7} M C l^{-}

4

1 0^{- 5} M  A g^{+} and 1 0^{- 5} M C l^{-}

Explanation:

$K_{S P} = [{Ag}^{+}] [{Cl}^{-}]$
to get ppt Ionic product $> K_{S P}$
$1.8 \times 10^{- 10} = 10^{- 4} \times 10^{- 4} = 10^{- 8}$
$10^{- 8} > 1.8 \times 10^{- 10}$ .

CHXI07:EQUILIBRIUM

314896 The mass of ${Pb}^{2 +}$ ion is left in solution when $50 mL of 0 .2 mL of 0 .2 M Pb(N O_{3})_{2}$ is added to $50 mL$ of 1.5 M NaCl is (Given $K_{SP}$ for $PbC l_{2}$ = $1 .7 \times 1 0^{- 4} M^{3 -}$

1

12 g

2

12 mg

3

120 g

4

120 mg

Explanation:

Assume complete precipitation of the ${Pb}^{2 +}$
followed by solution of the equilibrium
concentration to be determined.
Number of millimoles of
${Pb}^{2 +} = (50 mL) (0.2 mmol / mL) = 10 mmol$
Number of millimoles of
${Cl}^{-} = (50 mL) (1.5 mmol / mL) = 75 mmol$
${Pb}^{2 +} + 2 {Cl}^{-} \to {PbCl}_{2}$
What is the lead concentration
that can exist in a solution of
$C l^{-}$ containing $75 mol - 20mmol$
$= 55 m mol$ in $100 mL$
Concentration of
$C l^{-} = 55 m mol 100 mL = 0 . 55 M$
${PbCl}_{2} ⇌ {Pb}^{2 +} + 2 {Cl}^{-}$
$K_{s p}$ = $P b^{2 +} {[C l^{-}]}^{2} = 1 .7 1 0^{- 4} M^{3}$
$[P b^{2 +}] = \frac{1.7 10^{- 4}}{(0.55)^{2}} = 5.6 \times 10^{- 4} M$
$\frac{w_{2} \times 1000}{M W_{2} \times V (m L)} = 5.6 \times 10^{- 4}$
$\frac{w_{2} \times 1000}{208 \times 100} = 5.6 \times 10^{- 4}$ weight of ${Pb}^{2 +}$ ion $= 12 mg$

CHXI07:EQUILIBRIUM

314897 In the third group of qualitative analysis, the precipitating reagent is ${NH}_{4} Cl + {NH}_{4} OH$ . The function of ${NH}_{4} Cl$ is to

1 Increase the ionization of

{NH}_{4} OH

2 Supress the ionization of

{NH}_{4} OH

3 Stabilise the hydroxides of group cations

4 Convert the ions of group III into their respective chlorides

Explanation:

Due to common ion effect, ${NH}_{4} OH$ ionization is suppressed.

CHXI07:EQUILIBRIUM

314894 An aqueous solution contains an unknown concentration of ${Ba}^{2 +}$ . When $50 ml$ of a $1 M$ solution of ${Na}_{2} {SO}_{4}$ is added, ${BaSO}_{4}$ just begins to precipitate. The final volume is $500 mL$ . The solubility product of ${BaSO}_{4}$ is $1 \times 10^{- 10}$ . What is the original concentration of ${Ba}^{2 +}$ ?

1

2 \times 1 0^{- 9} M

2

1 .1 \times 1 0^{- 9} M

3

1 .0 \times 1 0^{- 10} M

4

5 \times 1 0^{- 9} M

Explanation:

$50 m L$ $1 M$ $N a_{2} S O_{4}$ is made to $500 mL$
means ${BaSO}_{4}$ solution is of $450 ml$ .
Concentration of ${SO}_{4}^{- 2}$ in ${Ba}^{+ 2}$ solution
$M_{1} V_{1} = M_{2} V_{2} 1 \times 50 = M_{2} \times 500 M_{2} = \frac{1}{10}$
for just precipitation
Ionic product $= K_{S P}$
$[B a^{+ 2}] [{SO}_{4}^{- 2}] = K_{SP} (BaS O_{4}) [B a^{+ 2}] \times \frac{1}{10} = 1 0^{- 10}$
$[B a^{+ 2}] = 1 0^{- 9} M in 500 ml solution$
for calculation of $[B a^{+ 2}]$ in original solution (450ml)
$M_{1} \times 450 = 10^{- 9} \times 500 M_{1} = \frac{500}{450} \times 10^{- 9} = 1.11 \times 10^{- 9} M$
$[M_{1} =$ molarity of ${Ba}^{+ 2}$ in original solution (450 $ml)$ ]

CHXI07:EQUILIBRIUM

314895 When equal volumes of the following solutions are mixed, precipitation of $AgCl (K_{sp} = 1.8 \times 10^{- 10})$ will only with

1

1 0^{- 4} M A g^{+} and 1 0^{- 4} M C l^{-}

2

1 0^{- 6} M A g^{+} and 1 0^{- 6} M C l^{-}

3

1 0^{- 7} M  A g^{+} and 1 0^{- 7} M C l^{-}

4

1 0^{- 5} M  A g^{+} and 1 0^{- 5} M C l^{-}

Explanation:

$K_{S P} = [{Ag}^{+}] [{Cl}^{-}]$
to get ppt Ionic product $> K_{S P}$
$1.8 \times 10^{- 10} = 10^{- 4} \times 10^{- 4} = 10^{- 8}$
$10^{- 8} > 1.8 \times 10^{- 10}$ .

CHXI07:EQUILIBRIUM

314896 The mass of ${Pb}^{2 +}$ ion is left in solution when $50 mL of 0 .2 mL of 0 .2 M Pb(N O_{3})_{2}$ is added to $50 mL$ of 1.5 M NaCl is (Given $K_{SP}$ for $PbC l_{2}$ = $1 .7 \times 1 0^{- 4} M^{3 -}$

1

12 g

2

12 mg

3

120 g

4

120 mg

Explanation:

Assume complete precipitation of the ${Pb}^{2 +}$
followed by solution of the equilibrium
concentration to be determined.
Number of millimoles of
${Pb}^{2 +} = (50 mL) (0.2 mmol / mL) = 10 mmol$
Number of millimoles of
${Cl}^{-} = (50 mL) (1.5 mmol / mL) = 75 mmol$
${Pb}^{2 +} + 2 {Cl}^{-} \to {PbCl}_{2}$
What is the lead concentration
that can exist in a solution of
$C l^{-}$ containing $75 mol - 20mmol$
$= 55 m mol$ in $100 mL$
Concentration of
$C l^{-} = 55 m mol 100 mL = 0 . 55 M$
${PbCl}_{2} ⇌ {Pb}^{2 +} + 2 {Cl}^{-}$
$K_{s p}$ = $P b^{2 +} {[C l^{-}]}^{2} = 1 .7 1 0^{- 4} M^{3}$
$[P b^{2 +}] = \frac{1.7 10^{- 4}}{(0.55)^{2}} = 5.6 \times 10^{- 4} M$
$\frac{w_{2} \times 1000}{M W_{2} \times V (m L)} = 5.6 \times 10^{- 4}$
$\frac{w_{2} \times 1000}{208 \times 100} = 5.6 \times 10^{- 4}$ weight of ${Pb}^{2 +}$ ion $= 12 mg$

CHXI07:EQUILIBRIUM

314897 In the third group of qualitative analysis, the precipitating reagent is ${NH}_{4} Cl + {NH}_{4} OH$ . The function of ${NH}_{4} Cl$ is to

1 Increase the ionization of

{NH}_{4} OH

2 Supress the ionization of

{NH}_{4} OH

3 Stabilise the hydroxides of group cations

4 Convert the ions of group III into their respective chlorides

Explanation:

Due to common ion effect, ${NH}_{4} OH$ ionization is suppressed.

CHXI07:EQUILIBRIUM

314894 An aqueous solution contains an unknown concentration of ${Ba}^{2 +}$ . When $50 ml$ of a $1 M$ solution of ${Na}_{2} {SO}_{4}$ is added, ${BaSO}_{4}$ just begins to precipitate. The final volume is $500 mL$ . The solubility product of ${BaSO}_{4}$ is $1 \times 10^{- 10}$ . What is the original concentration of ${Ba}^{2 +}$ ?

1

2 \times 1 0^{- 9} M

2

1 .1 \times 1 0^{- 9} M

3

1 .0 \times 1 0^{- 10} M

4

5 \times 1 0^{- 9} M

Explanation:

$50 m L$ $1 M$ $N a_{2} S O_{4}$ is made to $500 mL$
means ${BaSO}_{4}$ solution is of $450 ml$ .
Concentration of ${SO}_{4}^{- 2}$ in ${Ba}^{+ 2}$ solution
$M_{1} V_{1} = M_{2} V_{2} 1 \times 50 = M_{2} \times 500 M_{2} = \frac{1}{10}$
for just precipitation
Ionic product $= K_{S P}$
$[B a^{+ 2}] [{SO}_{4}^{- 2}] = K_{SP} (BaS O_{4}) [B a^{+ 2}] \times \frac{1}{10} = 1 0^{- 10}$
$[B a^{+ 2}] = 1 0^{- 9} M in 500 ml solution$
for calculation of $[B a^{+ 2}]$ in original solution (450ml)
$M_{1} \times 450 = 10^{- 9} \times 500 M_{1} = \frac{500}{450} \times 10^{- 9} = 1.11 \times 10^{- 9} M$
$[M_{1} =$ molarity of ${Ba}^{+ 2}$ in original solution (450 $ml)$ ]

CHXI07:EQUILIBRIUM

314895 When equal volumes of the following solutions are mixed, precipitation of $AgCl (K_{sp} = 1.8 \times 10^{- 10})$ will only with

1

1 0^{- 4} M A g^{+} and 1 0^{- 4} M C l^{-}

2

1 0^{- 6} M A g^{+} and 1 0^{- 6} M C l^{-}

3

1 0^{- 7} M  A g^{+} and 1 0^{- 7} M C l^{-}

4

1 0^{- 5} M  A g^{+} and 1 0^{- 5} M C l^{-}

Explanation:

$K_{S P} = [{Ag}^{+}] [{Cl}^{-}]$
to get ppt Ionic product $> K_{S P}$
$1.8 \times 10^{- 10} = 10^{- 4} \times 10^{- 4} = 10^{- 8}$
$10^{- 8} > 1.8 \times 10^{- 10}$ .

CHXI07:EQUILIBRIUM

314896 The mass of ${Pb}^{2 +}$ ion is left in solution when $50 mL of 0 .2 mL of 0 .2 M Pb(N O_{3})_{2}$ is added to $50 mL$ of 1.5 M NaCl is (Given $K_{SP}$ for $PbC l_{2}$ = $1 .7 \times 1 0^{- 4} M^{3 -}$

1

12 g

2

12 mg

3

120 g

4

120 mg

Explanation:

Assume complete precipitation of the ${Pb}^{2 +}$
followed by solution of the equilibrium
concentration to be determined.
Number of millimoles of
${Pb}^{2 +} = (50 mL) (0.2 mmol / mL) = 10 mmol$
Number of millimoles of
${Cl}^{-} = (50 mL) (1.5 mmol / mL) = 75 mmol$
${Pb}^{2 +} + 2 {Cl}^{-} \to {PbCl}_{2}$
What is the lead concentration
that can exist in a solution of
$C l^{-}$ containing $75 mol - 20mmol$
$= 55 m mol$ in $100 mL$
Concentration of
$C l^{-} = 55 m mol 100 mL = 0 . 55 M$
${PbCl}_{2} ⇌ {Pb}^{2 +} + 2 {Cl}^{-}$
$K_{s p}$ = $P b^{2 +} {[C l^{-}]}^{2} = 1 .7 1 0^{- 4} M^{3}$
$[P b^{2 +}] = \frac{1.7 10^{- 4}}{(0.55)^{2}} = 5.6 \times 10^{- 4} M$
$\frac{w_{2} \times 1000}{M W_{2} \times V (m L)} = 5.6 \times 10^{- 4}$
$\frac{w_{2} \times 1000}{208 \times 100} = 5.6 \times 10^{- 4}$ weight of ${Pb}^{2 +}$ ion $= 12 mg$

CHXI07:EQUILIBRIUM

314897 In the third group of qualitative analysis, the precipitating reagent is ${NH}_{4} Cl + {NH}_{4} OH$ . The function of ${NH}_{4} Cl$ is to

1 Increase the ionization of

{NH}_{4} OH

2 Supress the ionization of

{NH}_{4} OH

3 Stabilise the hydroxides of group cations

4 Convert the ions of group III into their respective chlorides

Explanation:

Due to common ion effect, ${NH}_{4} OH$ ionization is suppressed.