Question

Answer 1

Ta có: \(n_{HCl}=0,25.0,08=0,02\left(mol\right)\)

\(n_{H_2SO_4}=0,25.0,01=0,0025\left(mol\right)\)

\(\Rightarrow n_{H^+}=0,02+0,0025.2=0,025\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

⇒ n_OH^- (pư) = n_H+ = 0,025 (mol)

pH = 12 \(\Rightarrow\left[OH^-\right]=\dfrac{10^{-14}}{10^{-12}}=0,01\left(M\right)\)

\(\Rightarrow n_{OH^-\left(dư\right)}=0,01.0,5=0,005\left(mol\right)\)

\(\Rightarrow n_{Ba\left(OH\right)_2}=\dfrac{n_{OH^-}}{2}=\dfrac{0,025+0,005}{2}=0,015\left(mol\right)\)

\(\Rightarrow x=C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,015}{0,25}=0,06\left(M\right)\)

Có: \(n_{SO_4^{2-}}=n_{H_2SO_4}=0,0025\left(mol\right)\)

\(n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,015\left(mol\right)\)

PT: \(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_{4\downarrow}\)

\(\Rightarrow n_{BaSO_4}=0,0025\left(mol\right)\)

\(\Rightarrow m=0,0025.233=0,5825\left(g\right)\)

→ Đáp án: A