You visited us 1 times! Enjoying our articles? Unlock Full Access!

Rate of Reaction

Calculate the...

Question

Calculate the hydrolysis constant of analinium acetate if $K_{a} = 1 \times 10^{- 5}$ and $K_{b} = 1 \times 10^{10}$ .

1.24

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

10

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

0.1

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

0.01

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

Open in App

Solution

The correct option is C $0.1$
$H = 8.26$
$O H = 14 - p_{H}$
$= 14 - 8.26$
$= 5.74$
$k_{p} = 14 - 9.26$
$= 4.74$
$p O H = p k_{p} + log \frac{[N H_{4}^{+}]}{[N H_{4} O H]}$
$5.74 = 4.74 + log \frac{[{N H}_{4}^{+}]}{0.01}$
$1 = log [{N H}_{4}^{+}] - log (0.01)$
$1 = log [{N H}_{4}^{+}] + 2$
$log [{N H}_{4}^{+}] = - 1$
Taing antilog $[{N H}_{4}^{+}] = 0.1$

Suggest Corrections

Similar questions

(K_{h})

K_{a} =

1.8 \times

10^{- 5}

K_{b} =

2 \times

10^{- 5}

K_{w} = 10^{- 14}

Calculate the degree of hydrolysis ammonium acetate. Given $K_{a}$ = $1.75 \times 10^{- 5}$ ; $K_{b}$ = $1.80 \times 10^{- 5}$ .

C H_{3} C O_{2} N a

K_{a}

C H_{3} C O_{2} H = 1.8 \times 10^{- 5}

(h)

25^{\circ} C

K_{a}

C H_{3} C O O H

1.8 \times 10^{- 5}

K_{b}

4.6 \times 10^{- 10}

(K_{a} = 2 \times 10^{- 6})

(K_{b} = 5 \times 10^{- 7})

Join BYJU'S Learning Program