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Acidic Buffer Action

A student pre...

Question

A student prepares a buffer solution which contains $0.05 mol d m^{- 3}$ $H C O O H$ and $0.250 m o l d m^{- 3}$ $H C O O^{-} N a^{+} .$ The $K_{a}$ of $H C O O H$ is $1.78 \times 10^{- 4} .$ Calculate the pH of the solution.
$l o g 1.78 = 0.250, l o g 5 = 0.699$

12.5

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4.45

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6.2

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7.5

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Solution

The correct option is B 4.45
Given, $[H C O O H] = 0.05 mol d m^{- 3}, [H C O O^{-} N a^{+}] = 0.250 mol d m^{- 3}, K_{a} (H C O O H) = 1.78 \times 10^{- 4}$
Using the Henderson-Hasselbalch equation for weak bases:
$p H = p K_{a} + l o g [\frac{[c o n j u g a t e b a s e]}{[a c i d]}]$
$p H = p K_{a} + l o g [\frac{[H C O O^{-} N a^{+}]]}{[H C O O H]}]$
$p H = - l o g K_{a} + l o g [\frac{[H C O O^{-} N a^{+}]}{[H C O O H]}]$
$p H = - l o g K_{a} = + l o g [\frac{[H C O O^{-} N a^{+}]}{[H C O O H}]$
$p H = - l o g (1.78 \times 10^{- 4}) + l o g [\frac{[0.250]}{[0.05]}]$
$p H = 4 - l o g (1.78) + l o g (5)$
$p H = 4 - 0.250 + 0.699$
$p H = 4.45$

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Acidic Buffer Action

Standard XII Chemistry

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A student prepares a buffer solution which contains 0.05 mol dm−3 HCOOH and 0.250 mol dm−3 HCOO−Na+. The Ka of HCOOH is 1.78×10−4. Calculate the pH of the solution. log 1.78=0.250, log 5=0.699

A student prepares a buffer solution which contains $0.05 mol d m^{- 3}$ $H C O O H$ and $0.250 m o l d m^{- 3}$ $H C O O^{-} N a^{+} .$ The $K_{a}$ of $H C O O H$ is $1.78 \times 10^{- 4} .$ Calculate the pH of the solution.
$l o g 1.78 = 0.250, l o g 5 = 0.699$