Câu 1:

PT ion: \(H^++OH^-\rightarrow H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{OH^-}=0,6\cdot0,4+0,6\cdot0,3\cdot2=0,6\left(mol\right)\\n_{H^+}=0,2\cdot2,6=0,52\left(mol\right)\end{matrix}\right.\) 

\(\Rightarrow\) H⁺ hết, OH^- còn dư \(\Rightarrow n_{OH^-\left(dư\right)}=0,08\left(mol\right)\)

\(\Rightarrow\left[OH^-\right]=\dfrac{0,08}{0,6+0,2}=0,1\left(M\right)\) \(\Rightarrow pH=14+log\left(0,1\right)=13\)

Bài 2:

PT ion: \(H^++OH^-\rightarrow H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{OH^-}=0,3\cdot1,6=0,48\left(mol\right)\\n_{H^+}=0,2\cdot1\cdot2+0,2\cdot2=0,8\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\) OH^- hết, H⁺ còn dư \(\Rightarrow n_{H^+\left(dư\right)}=0,32\left(mol\right)\)

\(\Rightarrow\left[H^+\right]=\dfrac{0,32}{0,2+0,3}=0,64\left(M\right)\) \(\Rightarrow pH=-log\left(0,64\right)\approx0,19\)

$Ba(OH)_2 + 2HNO_3 \to Ba(NO_3)_2 + 2H_2O$
$V_{dd\ sau\ pư} = 0,2 + 0,3 = 0,5(lít)$
$n_{Ba(NO_3)_2} = n_{Ba(OH)_2} = 0,2.0,2 = 0,04(mol)$
$C_{M_{Ba(NO_3)_2}} = \dfrac{0,04}{0,5} = 0,08M$

a, \(\left\{{}\begin{matrix}n_{Ba^{2+}}=4.10^{-3}\left(mol\right)\\n_{Na^+}=3.10^{-3}\left(mol\right)\\n_{OH^-}=0,011\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[Ba^{2+}\right]=\dfrac{4.10^{-3}}{0,2+0,3}=0,008M\\\left[Na^+\right]=\dfrac{3.10^{-3}}{0,2+0,3}=0,006M\\\left[OH^-\right]=\dfrac{0,011}{0,2+0,3}=0,022M\end{matrix}\right.\)

b, Để trung hòa dung dịch A thì:

\(n_{H^+}=n_{OH^-}\)

\(\Leftrightarrow0,01.V_{ddHCl}=\left(0,02.2+0,01\right).0,2\)

\(\Leftrightarrow V_{ddHCl}=1\left(l\right)\)

cần lời giải chi tiết ạ

\(n_{NaOH}=0,2.1=0,2\left(mol\right)\\ n_{H_2SO_4}=0,3.1,5=0,45\left(mol\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

0,2------->0,1--------->0,1

Xét \(\dfrac{0,2}{2}< \dfrac{0,45}{1}\Rightarrow\) \(H_2SO_4\)dư

Trong dung dịch D có:

\(\left\{{}\begin{matrix}n_{H_2SO_4}=0,45-0,1=0,35\left(mol\right)\\n_{Na_2SO_4}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}CM_{H_2SO_4}=\dfrac{0,35}{0,5}=0,7M\\CM_{Na_2SO_4}=\dfrac{0,1}{0,5}=0,2M\end{matrix}\right.\)

b

\(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)

0,35<---------0,35

\(V_{Ca\left(OH\right)_2}=\dfrac{0,35.74}{1,2}=\dfrac{259}{12}\approx21,58\left(ml\right)\\ \Rightarrow V_{dd.Ca\left(OH\right)_2}=\dfrac{\dfrac{259}{12}.100\%}{10\%}=\dfrac{1295}{6}\approx215,83\left(ml\right)\)

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

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

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

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