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CHXI07:EQUILIBRIUM

314510 Assertion :
$pH$ of blood is maintained inspite of acidic foods.
Reason :
Acidity of foods is not so large to change the $pH$ of blood.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Dissolved ratio of
${HCO}_{3}^{-}$ to $H_{2} {CO}_{3}$ (buffer $H_{2} {CO}_{3} / {HCO}_{3}^{-}$ )
is helpful in controlling the $pH$ of the blood. So option (3) is correct.

CHXI07:EQUILIBRIUM

314511 The $pH$ of blood is 7.4. Assuming that the buffer in blood is carbon dioxide, hydrogen carbonate ion, what is the ratio of conjugate base to acid necessary to maintain blood at its proper $pH ? (Ka = 4.5 \times 10^{- 7})$

1 9

2 10

3 11

4 12

Explanation:

${CO}_{2} + 2 H_{2} O ⇌ H_{3} O^{+} + {HCO}_{3}^{-}$
${Ka}_{1} = \frac{[H_{3} O^{+}] [{HCO}_{3}^{-}]}{[{CO}_{3}^{2 -}]} = 4.5 \times 10^{- 7}$
Given that $pH = 7.4$ then $[H_{3} O^{+}]$ or $[H^{+}] = 4 \times 10^{- 8}$
$\frac{[{HCO}_{3}^{-}]}{[{CO}_{2}]} = \frac{4.5 \times 10^{- 7}}{4 \times 10^{- 8}} = 11$

CHXI07:EQUILIBRIUM

314512 The $pH$ of an acidic buffer mixture is

1 6.8

2 7

3 7.5

4 depends upon

K_{a}

of the acid

Explanation:

The $pH$ of acidic buffer is calculated by following formula
$pH = - \log K_{a} + \log [\frac{salt}{acid}]$
$∴ pH$ of acidic buffer depends on value of $K_{a}$ and is always less than 7 .

CHXI07:EQUILIBRIUM

314513 Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of $pH$ $6$ . If ionisation constant of HA is $10^{- 5}$ , the ratio of salt to acid concentration in the buffer solution will be

1

4 : 5

2

1 : 10

3

10 : 1

4

5 : 4

Explanation:

$HA ⇌ H^{+} + A^{-}$
(Unionised, weak acid and common ion effect)
$\begin{matrix} HA + NaOH ⟶ NaA + H_{2} O \\ NaA ⟶ {Na}^{+} + A^{-} (ionised) \\ Ka = \frac{[H^{+}] [A^{-}]}{[HA]} \end{matrix}$
Given, $p H = 6, [H^{+}] = 1 \times 10^{- 6}$
$[H^{+}] = \frac{K_{a} [Acid]}{[Salt]} \frac{[Salt]}{[Acid]} = \frac{K_{a}}{[H^{+}]} = \frac{10^{- 5}}{10^{- 6}} = 10 : 1$

JEE - 2017

CHXI07:EQUILIBRIUM

314510 Assertion :
$pH$ of blood is maintained inspite of acidic foods.
Reason :
Acidity of foods is not so large to change the $pH$ of blood.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Dissolved ratio of
${HCO}_{3}^{-}$ to $H_{2} {CO}_{3}$ (buffer $H_{2} {CO}_{3} / {HCO}_{3}^{-}$ )
is helpful in controlling the $pH$ of the blood. So option (3) is correct.

CHXI07:EQUILIBRIUM

314511 The $pH$ of blood is 7.4. Assuming that the buffer in blood is carbon dioxide, hydrogen carbonate ion, what is the ratio of conjugate base to acid necessary to maintain blood at its proper $pH ? (Ka = 4.5 \times 10^{- 7})$

1 9

2 10

3 11

4 12

Explanation:

${CO}_{2} + 2 H_{2} O ⇌ H_{3} O^{+} + {HCO}_{3}^{-}$
${Ka}_{1} = \frac{[H_{3} O^{+}] [{HCO}_{3}^{-}]}{[{CO}_{3}^{2 -}]} = 4.5 \times 10^{- 7}$
Given that $pH = 7.4$ then $[H_{3} O^{+}]$ or $[H^{+}] = 4 \times 10^{- 8}$
$\frac{[{HCO}_{3}^{-}]}{[{CO}_{2}]} = \frac{4.5 \times 10^{- 7}}{4 \times 10^{- 8}} = 11$

CHXI07:EQUILIBRIUM

314512 The $pH$ of an acidic buffer mixture is

1 6.8

2 7

3 7.5

4 depends upon

K_{a}

of the acid

Explanation:

The $pH$ of acidic buffer is calculated by following formula
$pH = - \log K_{a} + \log [\frac{salt}{acid}]$
$∴ pH$ of acidic buffer depends on value of $K_{a}$ and is always less than 7 .

CHXI07:EQUILIBRIUM

314513 Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of $pH$ $6$ . If ionisation constant of HA is $10^{- 5}$ , the ratio of salt to acid concentration in the buffer solution will be

1

4 : 5

2

1 : 10

3

10 : 1

4

5 : 4

Explanation:

$HA ⇌ H^{+} + A^{-}$
(Unionised, weak acid and common ion effect)
$\begin{matrix} HA + NaOH ⟶ NaA + H_{2} O \\ NaA ⟶ {Na}^{+} + A^{-} (ionised) \\ Ka = \frac{[H^{+}] [A^{-}]}{[HA]} \end{matrix}$
Given, $p H = 6, [H^{+}] = 1 \times 10^{- 6}$
$[H^{+}] = \frac{K_{a} [Acid]}{[Salt]} \frac{[Salt]}{[Acid]} = \frac{K_{a}}{[H^{+}]} = \frac{10^{- 5}}{10^{- 6}} = 10 : 1$

JEE - 2017

CHXI07:EQUILIBRIUM

314510 Assertion :
$pH$ of blood is maintained inspite of acidic foods.
Reason :
Acidity of foods is not so large to change the $pH$ of blood.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Dissolved ratio of
${HCO}_{3}^{-}$ to $H_{2} {CO}_{3}$ (buffer $H_{2} {CO}_{3} / {HCO}_{3}^{-}$ )
is helpful in controlling the $pH$ of the blood. So option (3) is correct.

CHXI07:EQUILIBRIUM

314511 The $pH$ of blood is 7.4. Assuming that the buffer in blood is carbon dioxide, hydrogen carbonate ion, what is the ratio of conjugate base to acid necessary to maintain blood at its proper $pH ? (Ka = 4.5 \times 10^{- 7})$

1 9

2 10

3 11

4 12

Explanation:

${CO}_{2} + 2 H_{2} O ⇌ H_{3} O^{+} + {HCO}_{3}^{-}$
${Ka}_{1} = \frac{[H_{3} O^{+}] [{HCO}_{3}^{-}]}{[{CO}_{3}^{2 -}]} = 4.5 \times 10^{- 7}$
Given that $pH = 7.4$ then $[H_{3} O^{+}]$ or $[H^{+}] = 4 \times 10^{- 8}$
$\frac{[{HCO}_{3}^{-}]}{[{CO}_{2}]} = \frac{4.5 \times 10^{- 7}}{4 \times 10^{- 8}} = 11$

CHXI07:EQUILIBRIUM

314512 The $pH$ of an acidic buffer mixture is

1 6.8

2 7

3 7.5

4 depends upon

K_{a}

of the acid

Explanation:

The $pH$ of acidic buffer is calculated by following formula
$pH = - \log K_{a} + \log [\frac{salt}{acid}]$
$∴ pH$ of acidic buffer depends on value of $K_{a}$ and is always less than 7 .

CHXI07:EQUILIBRIUM

314513 Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of $pH$ $6$ . If ionisation constant of HA is $10^{- 5}$ , the ratio of salt to acid concentration in the buffer solution will be

1

4 : 5

2

1 : 10

3

10 : 1

4

5 : 4

Explanation:

$HA ⇌ H^{+} + A^{-}$
(Unionised, weak acid and common ion effect)
$\begin{matrix} HA + NaOH ⟶ NaA + H_{2} O \\ NaA ⟶ {Na}^{+} + A^{-} (ionised) \\ Ka = \frac{[H^{+}] [A^{-}]}{[HA]} \end{matrix}$
Given, $p H = 6, [H^{+}] = 1 \times 10^{- 6}$
$[H^{+}] = \frac{K_{a} [Acid]}{[Salt]} \frac{[Salt]}{[Acid]} = \frac{K_{a}}{[H^{+}]} = \frac{10^{- 5}}{10^{- 6}} = 10 : 1$

JEE - 2017

CHXI07:EQUILIBRIUM

314510 Assertion :
$pH$ of blood is maintained inspite of acidic foods.
Reason :
Acidity of foods is not so large to change the $pH$ of blood.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Dissolved ratio of
${HCO}_{3}^{-}$ to $H_{2} {CO}_{3}$ (buffer $H_{2} {CO}_{3} / {HCO}_{3}^{-}$ )
is helpful in controlling the $pH$ of the blood. So option (3) is correct.

CHXI07:EQUILIBRIUM

314511 The $pH$ of blood is 7.4. Assuming that the buffer in blood is carbon dioxide, hydrogen carbonate ion, what is the ratio of conjugate base to acid necessary to maintain blood at its proper $pH ? (Ka = 4.5 \times 10^{- 7})$

1 9

2 10

3 11

4 12

Explanation:

${CO}_{2} + 2 H_{2} O ⇌ H_{3} O^{+} + {HCO}_{3}^{-}$
${Ka}_{1} = \frac{[H_{3} O^{+}] [{HCO}_{3}^{-}]}{[{CO}_{3}^{2 -}]} = 4.5 \times 10^{- 7}$
Given that $pH = 7.4$ then $[H_{3} O^{+}]$ or $[H^{+}] = 4 \times 10^{- 8}$
$\frac{[{HCO}_{3}^{-}]}{[{CO}_{2}]} = \frac{4.5 \times 10^{- 7}}{4 \times 10^{- 8}} = 11$

CHXI07:EQUILIBRIUM

314512 The $pH$ of an acidic buffer mixture is

1 6.8

2 7

3 7.5

4 depends upon

K_{a}

of the acid

Explanation:

The $pH$ of acidic buffer is calculated by following formula
$pH = - \log K_{a} + \log [\frac{salt}{acid}]$
$∴ pH$ of acidic buffer depends on value of $K_{a}$ and is always less than 7 .

CHXI07:EQUILIBRIUM

314513 Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of $pH$ $6$ . If ionisation constant of HA is $10^{- 5}$ , the ratio of salt to acid concentration in the buffer solution will be

1

4 : 5

2

1 : 10

3

10 : 1

4

5 : 4

Explanation:

$HA ⇌ H^{+} + A^{-}$
(Unionised, weak acid and common ion effect)
$\begin{matrix} HA + NaOH ⟶ NaA + H_{2} O \\ NaA ⟶ {Na}^{+} + A^{-} (ionised) \\ Ka = \frac{[H^{+}] [A^{-}]}{[HA]} \end{matrix}$
Given, $p H = 6, [H^{+}] = 1 \times 10^{- 6}$
$[H^{+}] = \frac{K_{a} [Acid]}{[Salt]} \frac{[Salt]}{[Acid]} = \frac{K_{a}}{[H^{+}]} = \frac{10^{- 5}}{10^{- 6}} = 10 : 1$

JEE - 2017