Đáp án:
5) $V_{Ba(OH)_2}= 1000\ (ml)$
6) $ a=0,249$
Giải thích các bước giải:
Bài 5/
PTHH: $OH^-+H^+ \to H_2O$
Ta có: $n_{H^+}=n_{HCl}=\dfrac{100.1,825\%}{36,5}=0,05\ (mol)$
$pH=13⇒ pOH=1⇒ [OH^-]=0,1⇒ C_M( Ba(OH)_2)=0,05\ mol$
$⇒ V_{Ba(OH)_2}=\dfrac{0,05}{0,05}=1\ (l)=1000\ (ml)$
Bài 6/
$n_{H^+}=0,2.0,5=0,1\ (mol)$
$pH=3⇒ H^+\ \text{dư}⇒ [H^+\ \text{dư}]=10^{-3}\ (M)$
$⇒ n_{H^+\ \text{dư}}=10^{-3}.0,4=4.10^{-4}\ (mol)$
$⇒ n_{OH^-\ \text{pư}} =0,1-4.10^{-4}=0,0996\ (mol)$
$⇒ n_{Ba(OH)_2}=0,0498\ (mol)$
$⇒ a=0,249$