nBa(OH)₂ = 0,002 mol; nKOH = 0,006 mol → ∑ n O H - = 0,002.2 + 0,006 = 0,01 mol

→  [OH^-] = n V = 0 , 01 0 , 2 + 0 , 2 = 0,025 →  pOH = –lg[OH^-] = -lg0,025 = 1,6

Mà pH + pOH = 14 →  pH = 12,4

Đáp án C

nBa(OH)₂ = 0,002 mol; nKOH = 0,006 mol → ∑ n O H - = 0,002.2 + 0,006 = 0,01 mol

→  [OH^-] = n V = 0 , 01 0 , 2 + 0 , 2 = 0,025 →  pOH = –lg[OH^-] = -lg0,025 = 1,6

Mà pH + pOH = 14 →  pH = 12,4

Đáp án C

Chọn C.

Chọn D

Em cần hỗ trợ câu nào? Em làm được câu nào chưa?

a, \(n_{H^+}=0,025.0,2=0,005\left(mol\right)\)

\(n_{OH^-}=0,01.2.0,3=0,006\left(mol\right)\)

\(\Rightarrow n_{OH^-dư}=0,001\left(mol\right)\)

\(\Rightarrow\left[OH^-\right]_{dư}=\dfrac{0,001}{1}=10^{-3}\)

\(\Rightarrow\left[H^+\right]=10^{-11}\)

\(\Rightarrow pH=11\)

b, \(n_{Fe^{2+}}=n_{SO_4^{2-}}=0,02.0,1=0,002\left(mol\right)\)

\(n_{Ba^{2+}}=0,01.0,3=0,003\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{BaSO_4\downarrow}=n_{SO_4^{2-}}=0,002\left(mol\right)\\n_{Fe\left(OH\right)_2\downarrow}=n_{OH^-dư}=0,001\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{\downarrow}=0,002.233+0,001.90=0,556\left(g\right)\)

\(n_{H^+}=n_{HCl}=0,5.0,3=0,15\left(mol\right)\\ n_{OH^-}=2.n_{Ba\left(OH\right)_2}=0,2.a.2=0,4a\left(mol\right)\\ Vì:pH=1\Rightarrow-log\left[H^+\right]=1\\ \Leftrightarrow\left[H^+\right]=0,1\left(M\right)\\ \Rightarrow\dfrac{n_{H^+\left(dư\right)}}{0,5}=0,1\\ \Rightarrow n_{H^+\left(dư\right)}=0,05\left(mol\right)\\ \Rightarrow0,15-0,4a=0,05\\ \Leftrightarrow a=0,25\)