The hydrogen ion concentration of 0.2M ${\mathrm{CH}}_3\mathrm{COOH}$ which is 4% ionized is ____N.

 It is given to find out the hydrogen ion concentration in N of 0.2M ${\mathrm{CH}}_3\mathrm{COOH}$ which is 4% ionized.

 $K_a=\frac{C\alpha \times C\alpha }{C(1-\alpha )}$

 $K_a=\frac{C\alpha \times C\alpha }{C(1-\alpha )}$

$K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$

$K_a=\frac{C\alpha \times C\alpha }{C(1-\alpha )}$

$K_a=\frac{C{\alpha }^2}{C(1-\alpha )}$

Since $\alpha$ is very small $1-\alpha \approx$ 1

Therefore, $K_a=C{\alpha }^2$

Therefore $\left[H^{+}\right]=C\alpha$

= $0.2\times \left(\frac{4}{100}\right)$

= $8.0\times {10}^{–3}N$

Hence, the answer is $8.0\times {10}^{–3}N$.