Calculate the pH of 0-1M solution of NH 4 OCN- K b for NH 3 is 1-75- 10 -5 and K a for HOCN is 3-3- 10 -4-

Ammonium acetate is a salt of weak acid and weak base.So pH=7+ 1 2 (p K _ a pK _ b ) pH=7+ 1 2 ( log K _ a +log K _ b ) #160; pH=7+ 1 ...

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Question

Calculate the pH of $0.1 M$ solution of ${N H}_{4} O C N . K_{b}$ for ${N H}_{3}$ is $1.75 \times 10^{- 5}$ and $K_{a}$ for $H O C N$ is $3.3 \times 10^{- 4}$ .

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[H^{+}]

0.1 M

H C O O H

0.1 M

H O C N

X \times 10^{- 3} M

K_{a}

H C O O H

H O C N

1.8 \times 10^{- 4}

3.3 \times 10^{- 4}

[H^{+}]

0.1

H C O O H

0.1

H O C N

K_{a}

H C O O H

H O C N

1.8 \times 10^{- 4}

3.3 \times 10^{- 4}

H C O O H

(K_{a} = 1.8 \times 10^{- 4})

H O C N

(K_{a} = 3.3 \times 10^{- 4})

[H_{3} O^{+}]

p H

0.01

N H_{4} C N

K_{a}

H C N = 6.2 \times 10^{- 10}

K_{b}

N H_{3} = 1.6 \times 10^{- 5}

0.003 M

N a O C N

K_{a}

H O C N = 3.33 \times 10^{- 4}

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