$20)$

Cho $V_{NaOH,0.05M}=V(ml)$

`->` $V_{NaOH,0.5M}=18-V(ml)$

$n_{NaOH,0.05M}=V \times C_M=0.05V(mmol)$

$n_{NaOH,0.5M}=V \times C_M=0.5(18-V)(mmol)$

$n_{NaOH,pha}=V \times C_M=18 \times 0.3(mmol)$

Bảo toàn số mol $NaOH:$

$n_{NaOH,0.05M}+n_{NaOH,0.5M}=n_{NaOH,pha}$

`<=>`$0.05V+0.5(18-V)=18 \times 0.3$

`=>`$V=V_{NaOH,0.05M}=8(mL)$

`=>`$V_{NaOH,0.5M}=18-8=10(mL)$

Do đó, cần thêm $8(mL)$ dung dịch $NaOH$ $0.05M$ vào $10(mL)$ dung dịch $NaOH$ $0.5M$ để thu được dung dịch $18(mL)$ $NaOH$ $0.3M.$