Đáp án:
Giải thích các bước giải:
$a)$
$HCl \to H^+ + Cl^-$
$[H^+] = C_{M_{HCl}} = 0,0015M$
$\to pH = -log([H^+]) = -log(0,0015) = 2,82$
$b)$
$NaOH \to Na^+ + OH^-$
$[OH^-] = C_{M_{NaOH}} = 0,0035M$
$\to [H^+] = \dfrac{10^{-14}}{[OH^-]}= 2,85.10^{-12}M$
$\to pH = -log(2,85.10^{-12}) = 11,54$
$c)$
$H_2SO_4 \to 2H^+ + SO_4^{2-}$
$[H^+] = 0,0045.2 = 0,009M$
$\to pH = -log(0,009) = 2,045$
$d)$
$Ba(OH)_2 \to Ba^{2+} + 2OH^-$
$[OH^-] = 0,02.2 = 0,04M$
$[H^+] = \dfrac{10^{-14}}{[OH^-]} = \dfrac{10^{-14}}{0,04} =2,5.10^{-13}$
$\to pH = -log(2,5.10^{-13}) = 12,6$
