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Byju's Answer

Standard XII

Chemistry

Salt of Weak Acid and Strong Base

K a for HCN...

Question

$K_{a}$ for $H C N$ is $5 \times 10^{- 10}$ at $25^{o} C$ . For maintaining a constant $p H$ of 9, the volume of $5 M K C N$ solution required to be added to $10 m L$ of $2 M H C N$ solution is

4 mL

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7.95 mL

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2 mL

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9.3 mL

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Solution

The correct option is C 2 mL
HCN and KCN will form buffer solution, therefore,

$p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]}$

$p H = - log 5 \times 10^{- 10} + log [\frac{5 \times V}{V + 10} / \frac{10 \times 2}{V + 10}]$

or $9 = - log 5 \times 10^{- 10} + log \frac{V}{4}$

$∴$ $V = 2 m L$

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Similar questions

K_{a}

for

H C N

5 \times 10^{- 10}

25^{\circ} C

. For maintaining a constant pH of

9

, the volume of

5 M K C N

solution required to be added to

10

mL of

2 M H C N

solution is:

[Use log

2 = 0.3

]

K_{a}

for

H C N

5 \times 10^{- 10}

25^{o} C

. For maintaining a constant

p H

9

, the volume of

5 M

K C N

solution required to be added to

10

ml of

2 M

H C N

solution is:

K_{a}

for

H C N

5 \times 10^{- 10}

25^{o} C

. For maintaining a constant

p H = 9

, the volume of

5 M

K C N

solution required to be added to

10 m l

2 M

H C N

solution is:[Given

{log}_{10} 5 = 0.69

: antilog

(- 0.3010) = 0.5

]

K_{a}

for HNC is

5 \times 10^{- 10}

at 25C for maintaining a constant pH of 9, the volume of 5 M KCN solution required to be added to 10 ml of 2M HCN solution is _________.

(log 2 = 0.3)

K_{a}

for HCN is

5 \times 10^{- 10}

25^{\circ} C

. For maintaining a constant pH of 9, the volume of

5 M K C N

solution required to be added to

10 m L

2 M H C N

solution is

(log 2 = 0.3 a n d l o g 5 = 0.69)

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Ka for HCN is 5×10−10 at 25oC. For maintaining a constant pH of 9, the volume of 5 M KCN solution required to be added to 10 mL of 2 M HCN solution is

The correct option is C 2 mLHCN and KCN will form buffer solution, therefore,pH=pKa+log[salt][acid]pH=−log5×10−10+log[5×VV+10/10×2V+10]or 9=−log5×10−10+logV4∴ V=2 mL

$K_{a}$ for $H C N$ is $5 \times 10^{- 10}$ at $25^{o} C$ . For maintaining a constant $p H$ of 9, the volume of $5 M K C N$ solution required to be added to $10 m L$ of $2 M H C N$ solution is