Câu 15:

`a)`

$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$

`C^o : 0,5 -> 0 -> 0`

`[]: 0,5 - x -> x -> x`

`=> (x^2)/(0,5-x) = 1,8.10^{-5}`

`=> x = 2,991.10^{-3}`

`=> pH = -log[H^+] = 2,52`

`b)`

`NaOH + CH_3COOH -> CH_3COONa + H_2O`

`n_{CH_3COOH} = 0,5.0,5 = 0,25 (mol)`

`pH = 3 => [H^+] = 10^{-3}`

Có: `K_a = ([CH_3COO^-][H^+])/([CH_3COOH])`

`=> ([CH_3COO^-])/([CH_3COOH]) = (1,8.10^{-5})/(10^{-3}) = 0,018`

`=> n_{CH_3COO^-} = (0,018)/(1 + 0,018) . 0,25 = 4,42.10^{-3} (mol)`

`=> n_{NaOH} = n_{CH_3COO^-} = 4,42.10^{-3} (mol)`

`=> m_{NaOH} = 4,42.10^{-3} . 40 = 0,1768(g)`