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Standard XII

Chemistry

Acidic Buffer Action

In one litre ...

Question

In one litre of 0.1 M acetic solution, one millimole of $H C l$ is added. The pH of the solution is:
$(K_{a} (C H_{3} C O O H) = 2 \times 10^{- 5})$ (log 2 = 0.3, log 3 = 0.5, log 5 = 0.7)

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Solution

$C H_{3} C O O H K_{a} = 2 \times 10^{- 5} ⇌ C H_{3} C O O^{-} + H^{+}$

$\begin{matrix} 0.1 M & 0 & 10^{- 3} M 0.1 M - x & x & (10^{- 3} + x) M \end{matrix}$
$K_{a} = 2 \times 10^{- 5} = \frac{x (10^{- 3} + x)}{0.1 - x}$
if we neglect $x$ wrt to $10^{- 3}$ then, $x = 2 \times 10^{- 3}$
So $x$ is not negligible wrt to $10^{- 3}$ but it is negligible wrt to 0.1
$2 \times 10^{- 5} = \frac{x (10^{- 3} + x)}{0.1} \Rightarrow x = 10^{- 3} M$
$[H^{+}] = 10^{- 3} + 10^{- 3} = 2 \times 10^{- 3} \Rightarrow p H = 2.7$

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In one litre of 0.1 M acetic solution, one millimole of HCl is added. The pH of the solution is: (Ka(CH3COOH)=2×10−5) (log 2 = 0.3, log 3 = 0.5, log 5 = 0.7)

CH3COOHKa=2×10−5⇌CH3COO−+H+ 0.1M 010−3M0.1M−x x(10−3+x) M Ka=2×10−5=x(10−3+x)0.1−x if we neglect x wrt to 10−3 then, x=2×10−3 So x is not negligible wrt to 10−3 but it is negligible wrt to 0.1 2×10−5=x(10−3+x)0.1⇒x=10−3M [H+]=10−3+10−3=2×10−3⇒pH=2.7

In one litre of 0.1 M acetic solution, one millimole of $H C l$ is added. The pH of the solution is:
$(K_{a} (C H_{3} C O O H) = 2 \times 10^{- 5})$ (log 2 = 0.3, log 3 = 0.5, log 5 = 0.7)