Calculate the...

Question

Calculate the $p H$ of $0.01$ M solution of $N H_{4} C N$ . The dissociation constants $K_{a}$ for $H C N = 6.2 \times 10^{- 10}$ and $K_{b}$ for $N H_{3} = 1.6 \times 10^{- 5}$ .

Open in App

Solution

$p H = 7 + \frac{1}{2 (p k a - p k b)}$
$p k a = - l o g k a$
$\Rightarrow - l o g (6.2 \times 10^{- 10})$
$\Rightarrow 10 - 0.79 = 9.21$
$p k b = - l o g k b$
$\Rightarrow - l o g (1.6 \times 10^{- 5})$
$\Rightarrow 5 - 0.20 = 4.8$
$7 + \frac{1}{2} [9.21 - 4.8]$
$7 + 1 / 2 [4.41] = 9.21$ .

Suggest Corrections

Similar questions

0.1 M

{N H}_{4} O C N . K_{b}

{N H}_{3}

1.75 \times 10^{- 5}

K_{a}

H O C N

3.3 \times 10^{- 4}

N H_{4} C N

(K_{a} = 6.2 \times 10^{- 10})

N H_{4} O H (K_{b} = 1.8 \times 10^{- 5})

N H_{4} C N

p H

0.02 M

(N H_{4} C N)

[K_{1}

H C N = 4.99 \times 10^{- 9}, K_{b}

N H_{4} O H = 1.77 \times 10^{- 5}]

[O H^{-}]

N H_{4} O H (k_{b}) = 1.6 \times 10^{- 5}

l o g 2 = 0.30

K_{a}

5 \times 10^{- 10}

25^{0} C