Giả sử: \(n_{NaOH}=x\left(mol\right)\)
Ta có: \(pH_{\left(HCl\right)}=3\Rightarrow\left[H^+\right]=10^{-3}M\Rightarrow n_{H^+}=10^{-3}.0,2=2.10^{-4}\left(mol\right)\)
PT ion: \(H^++OH^-\rightarrow H_2O\)
_____2.10 -4→2.10 -4________ (mol)
Dung dịch thu được có pH = 11 nên OH- dư.
⇒ [ H+ ] = 10 -11 M
⇒ [ OH- ](dư) = 10-3 M \(\Rightarrow n_{OH^-\left(dư\right)}=10^{-3}.0,2=2.10^{-4}\left(mol\right)\)
\(\Rightarrow x=\Sigma n_{OH^-}=n_{OH^-\left(pư\right)}+n_{OH^-\left(dư\right)}=4.10^{-4}\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,016\left(g\right)\)
Bạn tham khảo nhé!
