$a.$

$m_{dd,2}=15+a(gam)$

Ta có: $C_{1}\%m_{dd,1}=C_{2}\%m_{dd,2}$

`<=>`$98\% \times 15=20\% \times (15+a) $

`=>`$a=58.5(gam)$

$b.$

$n_{H_2SO_4}=\dfrac{5.88 \times 20}{100 \times 98}=0.012(mol)$

`=>`$n_{H^+}=0.012 \times 2 =0.024(mol)$

$pOH=14-pH=14-13=1$

`=>`$[OH^-]=10^{-1}(M)$

$n_{NaOH}=n_{OH^-}=10^{-1} \times 10^{-3}V=10^{-4}V(mol)$

Trung hòa hoàn toàn :

$n_{H^+}=n_{OH^-}$

`<=>`$0.024=10^{-4}V$

`=>`$V=240(mL)$

`a)`

`m_{H_2SO_4}=15.98\%=14,7(g)`

`m_{dd\ H_2SO_4\ sau}={14,7}/{20\%}=73,5(g)`

`->a=73,5-15=58,5(g)`

`b)`

`n_{H_2SO_4}={5,88.20\%}/{98}=0,012(mol)`

`H_2SO_4+2NaOH->Na_2SO_4+2H_2O`

`0,012->0,024(mol)`

`pH=13->[NaOH]=[OH^-]={10^{-14}}/{10^{-13}}=0,1M`

`->V_{dd\ NaOH}={0,024}/{0,1}=0,24(L)=240(mL)`