At 25oC- the dissociation constant of HCN and HF are 4-10-10 and 6-67-10-4 respectively- Calculate the pH of a mixture of 0-1 M HF and 0-1 M HCN-

HCN is a weak acid so H^+ due to it can be neglected as comparison to HF . [H^+]=√(KC)=√(6.7×10^ 4× 0.1)=√(67×10^ 6) =8.18×10^ 3 ...

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Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

At 25oC, th...

Question

At $25^{o} C$ , the dissociation constant of $H C N$ and $H F$ are $4 \times 10^{- 10}$ and $6.67 \times 10^{- 4}$ respectively. Calculate the $p H$ of a mixture of $0.1 M H F$ and $0.1 M$ $H C N$ .

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Solution

$H C N$ is a weak acid so $H^{+}$ due to it can be neglected as comparison to $H F$ .

$[H^{+}] = \sqrt{K C} = \sqrt{6.7 \times 10^{- 4} \times 0.1} = \sqrt{67 \times 10^{- 6}}$

$= 8.18 \times 10^{- 3}$

$p H = - log [8.18 \times 10^{- 3}] = 3 - log [8.18]$

$p H = 2.087$

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At 25oC, the dissociation constant of HCN and HF are 4×10−10 and 6.67×10−4 respectively. Calculate the pH of a mixture of 0.1 M HF and 0.1 M HCN.

HCN is a weak acid so H+ due to it can be neglected as comparison to HF.[H+]=√KC=√6.7×10−4×0.1=√67×10−6=8.18×10−3pH=−log[8.18×10−3]=3−log[8.18]pH=2.087

At $25^{o} C$ , the dissociation constant of $H C N$ and $H F$ are $4 \times 10^{- 10}$ and $6.67 \times 10^{- 4}$ respectively. Calculate the $p H$ of a mixture of $0.1 M H F$ and $0.1 M$ $H C N$ .

$H C N$ is a weak acid so $H^{+}$ due to it can be neglected as comparison to $H F$ .

$[H^{+}] = \sqrt{K C} = \sqrt{6.7 \times 10^{- 4} \times 0.1} = \sqrt{67 \times 10^{- 6}}$

$= 8.18 \times 10^{- 3}$

$p H = - log [8.18 \times 10^{- 3}] = 3 - log [8.18]$

$p H = 2.087$