Câu 4:

Đặt $V_{H_2SO_4}=x(l)$

$\to n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,3.0,2+2x.0,15=0,3x+0,06(mol)$

$V_B=x+0,3(l)$

$\to 0,3x+0,06=(x+0,3).0,25$

$\to x=0,3(l)$

Câu 5:

$n_{OH^-}=n_{KOH}+2n_{Ca(OH)_2}=0,15.0,6+0,25.x.2=0,5x+0,09(mol)$

$V_C=0,15+0,25=0,4l$

$\to n_{OH^-}=0,4.0,4=0,16(mol)$

$\to 0,5x+0,09=0,16$

$\to x=0,14(M)$

Câu 6:

Ta có: $V_1+V_2=600$                               $(1)$

$n_{H_2SO_4}=0,4.0,001V_1=4V_1.10^{-4}(mol)$

$n_{HNO_3}=0,3.0,001V_2=3V_2.10^{-4}(mol)$

$\to n_{H^+}=n_{HNO_3}+2n_{H_2SO_4}= 8V_1.10^{-4}+ 3V_2.10^{-4}(mol)$

$V_D=0,6l$

$[H^+]=0,5M$

$\to n_{H^+}=0,5.0,6=0,3(mol)$

$\to 8V_1.10^{-4}+3V_2.10^{-4}=0,3$   $(2)$

Từ $(1)(2)\to V_1=240(ml); V_2=360(ml)$

Câu 7:

Ta có: $V_1+V_2=480$                               $(1)$

$n_{KOH}=0,15.0,001V_1=1,5V_1.10^{-4}(mol)$

$n_{Ba(OH)_2}=0,25.0,001V_2=2,5V_2.10^{-4}(mol)$

$\to n_{OH^-}=n_{KOH}+2n_{Ba(OH)_2}=1,5V_1.10^{-4}+5V_2.10^{-4}(mol)$

$V_E=0,48l$

$[OH^-]=0,325M$

$\to n_{OH^-}=0,48.0,325=0,156(mol)$

$\to 1,5V_1.10^{-4}+5V_2.10^{-4}=0,156$   $(2)$

Từ $(1)(2)\to V_1=V_2=240(ml)$