Calculate the mass of NH 4 2 SO 4 in g that must be added to 500 mL of 0 . 2 M NH 3 to yield a solution of pH = 9 . 35 . K b for NH 3 = 1 . 78 × 10 - 5 .

Molarity of NH 3 Solution = Moles of NH 3 solution Volume of solution Moles of NH 3 = 0 . 2 × 0 . 5 = 0 . 1 pOH = 14 - pH pOH = 14 - 9 . 35 = 4 . 65 pK b = - log K b = - log 1 . 78 × 10 - 5 = 4 . 74 pOH = pK b - log base salt log base salt = 4 . 74 - 4 . 65 = 0 . 09 base salt = antilog 0 . 09 = 1 . 23 base = moles of NH 3 = 0 . 1 salt = 0 . 1 1 . 23 = 0 . 08 Moles of NH 4 + = 0 . 08 . NH 4 2 SO 4 ⇌ 2 NH 4 + + SO 4 2 - Moles of NH 4 2 SO 4 = moles of NH 4 + 2 = 0 . 04 . Amount of NH 4 2 SO 4 = 0 . 04 × 132 = 5 . 28 g .

A weak acid HA after treatment with 12 mL of 0 . 1 M strong base BOH has a pH of 5 . At the end point, the volume of same base required is 26 . 6 mL . Calculate K a of acid.

Acid is reacting with base as per question.So it is an example of neutralization reaction. For complete neutralization, Number of milliequivalents of acid = Number of milliequivalents of base Number of milliequivalents of acid = 26 . 6 × 0 . 1 = 2 . 66 Number of milliequivalents of base= 12 × 0 . 1 = 1 . 2 (Number of milliequivalents = N × V ) For partial neutralization, HA + BOH → BA + H 2 O Before neutralisation 2 . 66 1 . 2 0 0 After neutralisation 1 . 46 0 1 . 2 1 . 2 After the reaction, the resultant mixture has HA (acid) and BA (salt) and thus it can acts as buffer.It is an example of acidic buffer. P H = - logK a + log [ Salt ] [ Acid ] 5 = - logK a + log 1 . 2 1 . 46 5 = - logK a - 0 . 085 logK a = - 5 . 085 K a = antilog ( - 5 . 085 ) = 8 . 21 × 10 - 6

The solubility of Pb ( OH ) 2 in water is 6 . 7 × 10 - 6 M . Calculate the solubility of Pb ( OH ) 2 in a buffer solution of pH = 8 .

Solubility product constant ( K sp ) : It is the equilibrium constant for a solid substance dissolving in an aqueous solution. It is the measure of level at which solute dissolves in solution. Pb ( OH ) 2 → Pb 2 + + 2 OH - S 2 S As per the question, Solubility ( S ) of Pb ( OH ) 2 in water = 6 . 7 × 10 - 6 M . K sp = [ Pb 2 + ] [ OH - ] 2 = S × ( 2 S ) 2 = ( 6 . 7 × 10 - 6 ) × ( 2 × 6 . 7 × 10 - 6 ) 2 = 1 . 2 × 10 - 15 pOH = 14 - pH = 14 - 8 = 6 [ OH - ] = 10 - pOH = 10 - 6 M The solubility of Pb ( OH ) 2 in the buffer solution can be calculated as, K sp = [ Pb 2 + ] [ OH - ] 2 1 . 2 × 10 - 15 = [ Pb 2 + ] × ( 10 - 6 ) 2 [ Pb 2 + ] = 1 . 2 × 10 - 3 M Solubility of Pb ( OH ) 2 in water is: 1 . 2 × 10 - 3 M

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If you have saturated solution of ${CaF}_{2},$ then

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Important Questions on Ionic Equilibrium

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.4>Q 22

Which one is correct for the saturated solution of

{Ca}_{3} {({PO}_{4})}_{2}

salt if the

K_{sp}

2 .05 \times 10^{- 33}

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.4>Q 23

pH

of saturated solution of

Ba (OH)_{2}

12

, the value of

K_{sp}

is not:

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.5>Q 1

Calculate the mass of

{({NH}_{4})}_{2} {SO}_{4}

g

that must be added to

500 mL

0.2 M {NH}_{3}

to yield a solution of

pH = 9.35

K_{b}

for

{NH}_{3} = 1.78 \times 10^{- 5}

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IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.5>Q 2

40 mL

solution of weak base

BOH

is titrated with

0.1 N HCl

solution. The

pH

of solution is found to be

10.04

and

9.14

after the addition of

5.0 mL

and

20.0 mL

of acid respectively. Find out

K_{b}

for weak base.

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.5>Q 3

A weak acid

HA

after treatment with

12 mL

0.1 M

strong base

BOH

has a

pH

5

. At the end point, the volume of same base required is

26.6 mL

. Calculate

K_{a}

of acid.

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.5>Q 4

The solubility of

Pb (OH)_{2}

in water is

6.7 \times 10^{- 6} M

. Calculate the solubility of

Pb (OH)_{2}

in a buffer solution of

pH = 8

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JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.5>Q 5

The

pH

of a buffer is

4.745

. When

0.01

mole of

NaOH

is added to

1

litre of it, the

pH

changes to

4.832

, calculate its buffer capacity.

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IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.5>Q 6

What is the

pH

of the solution when

0.2

mole of hydrochloric acid is added to one litre of a solution containing

1 M

acetic acid and acetate ion? Assume that the total volume is one litre.

(K_{a} for {CH}_{3} COOH = 1.8 \times 10^{- 5})

If you have saturated solution of CaF2, then

Important Questions on Ionic Equilibrium

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If you have saturated solution of ${CaF}_{2},$ then