Phương trình ion rút gọn: Cu 2+ +OH- ->Cu(OH)2, tương ứng với phản ứng xảy ra trong

Đáp án A

PT:  Cu(NO₃)₂ + Ba(OH)₂ → Cu(OH)₂ + Na(NO₃)₂

$\begin{array}{l}C{u}^{2+}+2N{O}_3^{-}+B{a}^{2+}+2O{H}^{-}\to Cu{\left(OH\right)}_2+B{a}^{2+}+2N{O}_3^{-}\\ C{u}^{2+}+2O{H}^{-}\to Cu{\left(OH\right)}_2\end{array}$

CuSO₄ + Ba(OH)₂ → BaSO₄ + Cu(OH)₂

$\begin{array}{l}C{u}^{2+}+S{O}_4^{2-}+B{a}^{2+}+2O{H}^{-}\to Cu{\left(OH\right)}_2+BaS{O}_4\\ Cu\text{}S{O}_4+H_{2}S\to CuS+H_{2}S{O}_4\\ C{u}^{2+}+S{O}_4^{2-}+H_{2}S\to CuS+H_{2}S{O}_4\end{array}$