When 100 ml of 0.4 M `CH_3 COOH ` is mixed with 100 ml of 0.2 M NaOH , the `[H_3O]^(+) ` in the solution is approximately : ` [K_a (CH_3COOH)= 1.8 xx 10 ^(-5) ]`

Options

A. `1.8 xx 10 ^(-4) `
B. ` 1.8 xx 10 ^(-5)`
C. ` 9 xx 10 ^(-6) `
D. ` 9xx 10 ^(-5) `

Answer

B

Explanation

`[H^(+) ] =([CH_3COOH])/([CH_3COO^(-)]) `
` =1.8 xx 10 ^(-5) xx ( (100 xx 0.4- 100 xx 0.2))/( (100 xx 0.2 )) =1.8 xx 10 ^(_5)`