Đáp án:

Giải thích các bước giải:

Câu 16 :

$H_2SO_4 \to 2H^+ + SO_4^{2-}$

$Al_2(SO_4)_3 \to 2Al^{3+} + 3SO_4^{2-}$

Ta có :

$[H^+] = 2C_{M_{H_2SO_4}} = 0,2.2 = 0,4M$
$[Al^{3+}] = 2C_{M_{Al_2(SO_4)_3}} = 0,01.2 = 0,02M$

$[SO_4^{2-}] = C_{M_{H_2SO_4}} + 3C_{M_{Al_2(SO_4)_3}} = 0,2 + 0,01.3=0,23M$

Câu 17 :

$V_{dd} = 200 + 300 = 500(ml) = 0,5(lít)$

Sau khi trộn :
$C_{M_{H_2SO_4}} = \dfrac{0,2.0,1}{0,5} = 0,04M$

$C_{M_{Al_2(SO_4)_3}} = \dfrac{0,3.0,12}{0,5} = 0,072M$

Suy ra :

$[H^+] = 2C_{M_{H_2SO_4}} = 0,04.2 = 0,08M$
$[Al^{3+}] = 2C_{M_{Al_2(SO_4)_3}} = 0,072.2 = 0,144M$

$[SO_4^{2-}] = C_{M_{H_2SO_4}} + 3C_{M_{Al_2(SO_4)_3}} = 0,04 + 0,072.3=0,256M$

Câu 18 :

$n_{HCl} = \dfrac{336}{1000.22,4} = 0,015(mol)$
$\to [H^+] = C_{M_{HCl}} = \dfrac{0,015}{0,2} = 0,075M$
$\to pH = -log([H^+]) = -log(0,075) = 1,125$

Câu 19 :

$[OH^-] = C_{M_{NaOH}} + 2C_{M_{Ba(OH)_2}} = 0,05 + 0,01.2 = 0,07M$

$\to pOH = -log(0,07) = 1,155$

$\to pH = 14 -1,155 = 12,845$

Câu 16: 

$[H^+]=2C_{H_2SO_4}=0,4M$

$[Al^{3+}]=2C_{Al_2(SO_4)_3}=0,02M$

$[SO_4^{2-}]=C_{H_2SO_4}+3C_{Al_2(SO_4)_3}=0,23M$

Câu 17:

$n_{H_2SO_4}=0,2.0,1=0,02 mol$

$n_{Al_2(SO_4)_3}=0,3.0,12=0,036 mol$

Sau khi trộn, $V_{dd}=0,5l$

$C_{H_2SO_4}=\dfrac{0,02}{0,5}=0,04M$

$C_{Al_2(SO_4)_3}=\dfrac{0,036}{0,5}=0,072M$

Suy ra:

$[H^+]=0,04.2=0,08M$

$[SO_4^{2-}]=0,04+0,072.3=0,256M$

$[Al^{3+}]=0,072.2=0,144M$ 

Câu 18:

$n_{HCl}=\dfrac{0,336}{22,4}=0,015 mol$

$\Rightarrow [H^+]=C_{HCl}=\dfrac{0,015}{0,2}=0,075M$

$pH=-\lg0,075=1,12$ 

Câu 19:

$[OH^-]=C_{NaOH}+2C_{Ba(OH)_2}=0,05+2.0,01=0,07M$

$\Rightarrow pH=14-pH=14+\lg[OH^-]=14-12,85=1,15$