Question

Calculate the molar concentration of an acetic acid solution which is 2% ionized? (Ka = 1.8 × $10^{-5}$)

Answer 1

"Given, $K_{ a }\quad =\quad 1.8\quad \times\quad { 10 }^{ -5 }$  and acetic acid solution is 2% ionized.

$K_{a}\quad =\quad \frac { [H^{ + }]\quad [CH_{ 3 }COO^{ - }] }{ \quad [CH_{ 3 }COOH] }$

$1.8\quad \times \quad { 10 }^{-5}\quad =\quad \frac { 0.02x\quad +\quad 0.02 x }{ x }$

$0.0004\quad x^{ 2 }\quad =\quad \left( 1.8\quad \times\quad { 10 }^{ -5 } \right) x$

$0.0004\quad x^{ 2 }\quad -\quad 0.000015\quad x\quad =\quad 0$

Using the quadratic equation, $x\quad =\quad \frac { [-b\pm \sqrt { ({ b }^{ 2 }\quad -\quad 4ac) } ] }{ 2a }$

Here, a = 0.0004; b = -0.000015; c = 0  

Solving we get, x = 0.0375 M "