You visited us 2 times! Enjoying our articles? Unlock Full Access!

Auto Protolysis of Water

Find the pH...

Question

Find the $p H$ of the resulting solution and then mark the option in which $p H$ exists between color transition range of an indicator.

$50$ ml of $0.1$ M $H C O_{3}^{-} +$ $50$ ml of $0.8$ M $C O_{3}^{2 -}$ .
[For $H_{2} C O_{3}$ : $K_{a_{1}} = 4 \times 10^{- 7}$ and $K_{a_{2}} = 2 \times 10^{- 11}$ ]

Phenol red (6.8 to 8.4)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Propyl red 4(4.6 to 6.4)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Phenolphthalein (8.3 to 10.1)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Malachite green (11.4 to 13)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D Malachite green (11.4 to 13)
Given that
[Salt] = 0.8M and [Acid] = 0.1M

The solution can be regarded as acidic buffer solution containing weak acid sodium bicarbonate and its salt sodium carbonate with strong base.
The expression for the pH of the acidic buffer solution is as given below.
$p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]}$
$p K_{a} = - l o g K_{a} = - l o g [2 \times 10^{- 11}] = 10.7$
Substitute values in the above solution.
$p H = 10.7 + l o g \frac{0.8}{0.1} = 11.6$
Hence, the suitable indicator is Malachite green with pH range from 11.4 to 13.

Suggest Corrections

Similar questions

Δ p H

N a H C O_{3}

p H = 9, K a_{1} = 10^{- 7}, K a_{2} = 10^{- 11}

H_{2} C O_{3}

Δ p H

N a H C O_{3}

(i n i t i a l p H = 9, K a_{1} = 10^{- 7}, K a_{2} = 10^{- 11}

H_{2} C O_{3})

100 m L

0.1 M

H_{3} P O_{4}

200 m L

0.1 M

N a O H

p H

G i v e n K_{a 1} = 10^{- 3}, K_{a 2} = 10^{- 8}, K_{a 3} = 10^{- 13}

100 mL

0.1 M NaOH

100 mL

0.05 M H_{2} S O_{4}

p H

(for H_{2} S O_{4}, K a_{1} = \infty, K a_{2} = 10^{- 2})

p H

C o (O H)_{3}

p H

C o (O H)_{3}, K_{s p} = 2.7 \times 10^{- 43}

Join BYJU'S Learning Program