$V_1=V_{X}(L);V_2=V_Y(L)$

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

$0,02V_1$ `->` $0,04V_1(M)$

$NaOH \to OH^-+Na^+$

$0,035V_2$ `->` $0,035V_2(M)$

$pH=2>7$`->`Acid dư.

`=>`$[H^+]_{dư}=10^{-2}(M)$

`=>`$n_{H^+}=(V_1+V_2) \times 10^{-2}(mol)$$(1)$

$OH^-+OH^+ \to H_2O$

$0,035V_2$`->`$0,035V_2$

`=>`$n_{H^+,dư}=0,04V_1-0,035V_2(mol)$$(2)$

$(1),(2)$`=>`$(V_1+V_2) \times 10^{-2}=0,04V_1-0,035V_2$

`<=>`$10V_1+10V_2=40V_1-35V_2$

`<=>`$45V_2=30V_1$

`<=>`$\dfrac{V_1}{V_2}=\dfrac{3}{2}$

`n_{H^+}=2n_{H_2SO_4}=2.0,02.V_X=0,04V_X(mol)`

`n_{OH^-}=n_{NaOH}=0,035V_Y(mol)`

`H^{+}+OH^{-}->H_2O`

`0,035V_Y←0,035V_Y(mol)`

`pH=2->[H^+\ sau]=0,01M`

`->n_{H^+\ sau}=0,01(V_X+V_Y)(mol)`

`->0,01(V_X+V_Y)+0,035V_Y=0,04V_X`

`->{V_X}/{V_Y}={0,035+0,01}/{0,04-0,01}=3/2`