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Đáp án:

`1,2M`

`0,8M`

Giải thích các bước giải:

`n_{KOH}=0,035.2=0,07 mol`

`n_{HCl}=0,025C_1 mol`

`n_{H_2SO_4} =0,025C_2 mol`

`KOH + HCl \to KCl + H_2O `

`2KOH + H_2SO_4 \to K_2SO_4 + 2H_2O `

Theo pt:

`n_{HCl}+2.n_{H_2SO_4} =n_{KOH }`

`=> 0,025C_1+0,05C_2=0,07`       `(1)`

Mặt khác:

Theo pt:

`n_{KCl}=n_{HCl}=0,025C_1 mol`

`n_{K_2SO_4}=n_{H_2SO_4} =0,025C_2 mol`

Khối lượng muối là:

`m_M=74,5.0,025C_1+174.0,025C_2=5,715g`

`=> 1,8625C_1+4,35C_2=5,715`       `(2)`

$(1)(2)=>\begin{cases} C_1=1,2 \\ C_2=0,8 \end{cases}$

$n_{KOH}=0,035.2=0,07(mol)$

$n_{HCl}=0,025C_1(mol)$

$n_{H_2SO_4}=0,025C_2(mol)$

$KOH+HCl\to KCl+H_2O$

$2KOH+H_2SO_4\to K_2SO_4+2H_2O$

$\to 0,025C_1+0,025C_2.2=0,07$  $(1)$

$n_{KCl}=n_{HCl}=0,025C_1(mol)$

$n_{K_2SO_4}=n_{H_2SO_4}=0,025C_2(mol)$

$\to 0,025C_1.74,5+0,025C_2.174=5,715$  $(2)$

$(1)(2)\to C_1=1,2(M); C_2=0,8(M)$