\(n_{HCl}=0.1\cdot0.03=0.003\left(mol\right)\)

\(n_{NaOH}=0.1\cdot0.01=0.001\left(mol\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

Lập tỉ lệ : 

\(\dfrac{0.003}{1}>\dfrac{0.001}{1}\Rightarrow HCldư\)

\(n_{HCl\left(dư\right)}=0.003-0.001=0.002\left(mol\right)\)

\(\left[H^+\right]=\dfrac{0.002}{0.1+0.1}=0.01\)

\(pH=-log\left(0.01\right)=2\)

\(b.\)

\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

\(0.001..........0.002\)

\(V_{Ba\left(OH\right)_2}=\dfrac{0.001}{1}=0.001\left(l\right)\)

\(pH=7\Rightarrow n_{H^+}=n_{OH^-}\)

\(\Leftrightarrow\left(0,05+0,06.2\right)\text{​​}V_2=\left(0,08+0,02.2\right)V_1\)

\(\Rightarrow V_1:V_2=17:12\)

Sửa đề H₂SO₂ thành H₂SO₄

\(n_{Ba^{2+}}=n_{Ba\left(OH\right)2}=0,01.0,1=0,001\left(mol\right)\)

\(\Rightarrow n_{OH^-}=2n_{Ba\left(OH\right)2}=2.0,001=0,002\left(mol\right)\)

\(n_{SO_4^{2-}}=n_{H2SO4}=0,1.0,05=0,005\left(mol\right)\)

\(\Rightarrow n_{H^+}=2n_{H2SO4}=2.0,005=0,01\left(mol\right)\)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)

0,001   0,01           0,01

Xét tỉ lệ : \(0,001< 0,01\Rightarrow SO_4^{2-}dư\)

\(n_{Ba^{2+}\left(pư\right)}=n_{BaSO4}=0,001\left(mol\right)\Rightarrow m_{BaSO4}=0,001.233=0,233\left(g\right)\)

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

0,01 0,002

Xét tỉ lệ : \(0,01>0,002\Rightarrow H^+dư\)

\(n_{H^+dư}=0,01-0,002=0,008\left(mol\right)\Rightarrow\left[H^+\right]=\dfrac{0,008}{0,1+0,1}=0,04M\)

\(\Rightarrow pH=-log\left(0,04\right)\approx1,4\)

\(n_{OH^-}=0,12V_1\)

\(n_{H^+}=0,17V_2\)

\(n_{OH^-dư}=\left(V_1+V_2\right).10^{-1}\)

Ta có: 

\(n_{OH^-dư}+n_{H^+}=n_{OH^-}\)

\(\Leftrightarrow\left(V_1+V_2\right).10^{-1}=0,12V_1\)

\(\Leftrightarrow0,1V_1=0,02V_2\)

\(\Rightarrow\dfrac{V_1}{V_2}=\dfrac{1}{5}\)

$n_{NaOH} = 0,001(mol)$
$n_{Ba(OH)_2} = 0,01.0,15 = 0,0015(mol)$

$NaOH \to Na^+ + OH^-$
$Ba(OH)_2 \to Ba^{2+} + 2OH^-$

Ta có : 

$n_{OH^-}= 0,001 + 0,0015.2 = 0,004(mol)$
$V_{dd} = 0,1 + 0,15 = 0,25(mol)$
$[OH^-] = \dfrac{0,004}{0,25} = 0,016M$

$pOH = -log(0,016) = 1,795 \Rightarrow pH = 14 - 1,795 = 12,205$

\(n_{OH^-}=0,1.0,01+0,15.0,01.2=0,004\left(mol\right)\\ V_{ddsau}=100+150=250\left(ml\right)=0,25\left(l\right)\\ \Rightarrow\left[OH^-\right]=\dfrac{0,004}{0,25}=0,016\left(M\right)\\ pH=14+log\left[0,016\right]\approx12,204\)

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

PT ion: \(H^++OH^-\rightarrow H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{H^+}=0,15\cdot0,05=0,0075\left(mol\right)\\n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,001\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\) H⁺ còn dư 0,0065 mol

\(\Rightarrow\left[H^+\right]=\dfrac{0,0065}{0,2}=0,0325\left(M\right)\) \(\Rightarrow pH=-log\left(0,0325\right)\approx1,5\)