2 câu trả lời
* KOH:
Ta có:
$CM_{KOH}$ = 1 . $10^{-4}$ (M)
-> [OH-] = 1 . $10^{-4}$ (M)
-> pOH = -log[OH-] = -log(1 . $10^{-4}$) = 4
-> pH = 14 - 4 = 10.
* H2SO4:
Ta có:
$CM_{H2SO4}$ = 0,5 . $10^{-4}$ (M)
-> [H+] = (0,5 . $10^{-4}$).2 = $10^{-4}$ (M)
-> pH = -log[H+] = -log($10^{-4}$) = 4.
`KOH`:
`[OH^{-}] = 10^{-4}`
`pH = 14 + log(10^{-4}) = 10`
`H_2SO_4`:
`[H^{+}] = 0,5. 10^{-4} = 5.10^{-5}`
`pH = -log(5.10^{-5}) = 4,3`
