A 0.01 M acetic acid solution is 1% ionised. An another acetic acid is 10% ionised. What will be the concentration of another acetic acid?

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Answer 1 · 2018-01-07T01:03:53

The concentration will be $9 \times 10^{-5} l mol/L$ $9 × 10^"-5"color(white)(l) "mol/L"$ .

Let $c$ $c$ be the initial concentration of acetic acid.

The equation for the equilibrium is

$m HA + H_{2} O ⇌ A^{-} + H_{3} O^{+}$ $color(white)(m)"HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"$
$c-αc color(white)(mmmmm) αc color(white)(mml) αc$

The equilibrium concentration of $"HA"$ will decrease by $αc$ , and those of $"A"^"-"$ and $"H"_3"O"^"+"$ will be $αc$ .

$K = (αc)^2/(c-αc) = (α^2c^2)/(c(1-α)) = (α^2c)/(1-α)$

For the first solution

$K = ((0.01)^2× 0.01)/(1-0.01) = (1 × 10^"-6")/0.99 = 1.0 × 10^"-6"$

For the second solution,

$α^2c = K(1-α)$

$c= (K(1-α))/α^2$

$c = (1.0 × 10^"-6"(1-0.10))/(0.10)^2 = 9 × 10^"-5"color(white)(l) "mol/L"$