Calculate the pH at which an acid indicator with K a = 1 × 10 - 5 changes colour when the indicator concentration is 1 × 10 - 3 M . Also report the pH at which coloured ion is 80 % present. ( log 2 = 0 . 3 ) .

The acid indicator changes their colour at mid point of their range; when the concentration of acid and its conjugate base will be equal. Consider an acid Indicator HIn which dissociates in aqueous medium as follows HIn ⇌ H + ( aq ) + I n ( aq ) ⊝ Ka = H + I n - HI n Since I n - = HI n on color changing so K a = H + 1 × 10 - 5 = H + pH = - log H + = - log 10 - 5 pH = 5 When Indicator 80 % coloured of ion then in such case I n - ≠ HIn But the Intensity of colour of HIn decreases. HIn ⇌ H ( aq ) + + I n ⊝ ( aq ) 10 - 3 0 0 10 - 3 × 0 . 20 10 - 3 × 0 . 80 Here In - 1 is 80 % and HIn will 20 % Ka = H + 10 - 3 × 0 . 80 10 - 3 × 0 . 20 K a = 10 - 5 So H + = 1 4 × 10 - 5 pH = - log 10 - 5 × 1 4 pH = 5 . 6

Calculate the pH during the titration of 40 . 00 mL of 0 . 1 M propionic acid HPr ; K a = 1 × 10 - 5 after adding the following volumes of 0 . 1 M NaOH : (a) 0 . 0 ml Given, ( log 2 = 0 . 3 , log 3 = 0 . 48 )

HPr is weak acid here millimoles of HPr is = 40 × 0 . 1 = 4 millimoles Millimoles of NaOH = 0 . 1 × 0 = 0 millimoles So, pH will only due to HPr For weak acids H + = Ka × c = 10 - 5 × 0 . 1 = 10 - 6 = 10 - 3 Hence, pH = - log 10 - 3 = 3

Calculate the pH during the titration of 40 . 00 mL of 0 . 1 M propionic acid HPr ; K a = 1 × 10 - 5 after adding the following volumes of 0 . 1 M NaOH : ( log 2 = 0 . 3 , log 3 = 0 . 48 ) 30 . 00 mL

HPr + OH - → Pr - + H 2 O Initial 4 3 1 3 The solution is a buffer solution , so, pH = pK a + log [ Pr - ] [ HPr ] = 5 + log 3 = 5 + 0 . 48 = 5 . 48 .

Calculate the pH during the titration of 40 . 00 ml of 0 . 1 M propionic acid HPr ; K a = 1 × 10 - 5 , after adding the following volumes of 0 . 1 M NaOH : 40 . 00 ml ( log 2 = 0 . 3 , log 3 = 0 . 48 )

HPr is weak acid here millimoles of HPr is = 40 × 0 . 1 = 4 millimoles Millimoles of NaOH = 0 . 1 × 40 millimoles millimoles Therefore pH can be calculated by salt hydrolysis (salt of a weak acid with the strong base) The net concentration of the salt is C C = 40 × 0 . 1 40 + 40 = 4 80 = 1 20 M pK a = - log ( 10 - 5 ) = 5 Hence, pH = 7 + 1 2 pK a + logC = 7 + 1 2 × 5 + 1 2 log 0 . 05 = 8 . 85

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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 7.5>Q 14

Which of the following will function as buffer?

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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 15

In a buffer solution consisting of ${NaH}_{2} {PO}_{4}$ and ${Na}_{2} {HPO}_{4}$

$(1) {NaH}_{2} {PO}_{4}$ is a acid and ${Na}_{2} {HPO}_{4}$ is salt.

$(2)$ The $pH$ of the solution can be calculated using the relation: $pH = K_{a_{2}} + \log_{10} \frac{[{HPO}_{4}^{2 -}]}{H_{2} {PO}_{4}^{-}]}$ .

$(3)$ The ${Na}_{2} {HPO}_{4}$ is acid and ${NaH}_{2} {PO}_{4}$ is salt.

$(4)$ The $pH$ cannot be calculated.

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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 16

In a buffer solution of

{NaH}_{2} {PO}_{4}

and

{Na}_{2} {HPO}_{4}

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.6>Q 1

Calculate the

pH

at which an acid indicator with

K_{a} = 1 \times 10^{- 5}

changes colour when the indicator concentration is

1 \times 10^{- 3} M

. Also report the

pH

at which coloured ion is

80 %

present.

(\log 2 = 0.3)

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.6>Q 2

15 ml

sample of

0.15 M {NH}_{3} (aq)

is titrated against

15 ml of 0.1 M HCl (aq)

. What is the

pH

at the end point?

K_{b}

{NH}_{3} (aq) = 1.8 \times 10^{- 5}

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.6>Q 3

Calculate the $pH$ during the titration of $40.00 mL$ of $0.1 M$ propionic acid $(HPr; K_{a} = 1 \times 10^{- 5})$ after adding the following volumes of $0.1 M NaOH :$

(a) $0.0 ml$

Given, $(\log 2 = 0.3, \log 3 = 0.48)$

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.6>Q 3

Calculate the $pH$ during the titration of $40.00 mL$ of $0.1 M$ propionic acid $(HPr; K_{a} = 1 \times 10^{- 5})$ after adding the following volumes of $0.1 M NaOH : (\log 2 = 0.3, \log 3 = 0.48)$

$30.00 mL$

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.6>Q 3

Calculate the $pH$ during the titration of $40.00 ml$ of $0.1 M$ propionic acid $(HPr; K_{a} = 1 \times 10^{- 5})$ , after adding the following volumes of $0.1 M NaOH :$

$40.00 ml$

$(\log 2 = 0.3, \log 3 = 0.48)$

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Important Questions on Ionic Equilibrium

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