\(n_{NaOH}=0,1.0,01=0,001(mol)\\ \Rightarrow n_{OH^{-}}=0,001(mol)\\ n_{HCl}=0,03.0,2=0,006(mol)\\ \Rightarrow n_{H^{+}}=0,006(mol)\\ H^{+}+OH^{-}\to H_2O\\ 0,001<0,006\\ OH^{-} hêt; H^{+} dư\\ n_{H^{+}}=0,006-0,001=0,005(mol)\\ [H^{+}]=\frac{0,005}{0,1+0,2}=\frac{1}{60}M\\ \to pH=-log(\frac{1}{60})=1,77 \)

Ta có: \(\left\{{}\begin{matrix}n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,1.0,01=0,001\left(mol\right)\\n_{OH^-}=2n_{Ba\left(OH\right)_2}=2.0,1.0,01=0,002\left(mol\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\Sigma n_{H^+}=0,2.0,01+2.0,2.0,02=0,01\left(mol\right)\\n_{SO_4^{2-}}=n_{H_2SO_4}=0,2.0,02=0,004\left(mol\right)\end{matrix}\right.\)

PT ion: \(H^++OH^-\rightarrow H_2O\)

_____ 0,01___0,002_________ (mol)

⇒ H⁺ dư. \(\Rightarrow n_{H^+\left(dư\right)}=0,008\left(mol\right)\)

\(\Rightarrow\left[H^+\right]=\frac{0,008}{0,3}=\frac{2}{75}M\Rightarrow pH\approx1,57\)

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

______ 0,001__0,004__ → 0,001 (mol)

\(\Rightarrow m_{\downarrow}=m_{BaSO_4}=0,001.233=0,233\left(g\right)\)

Bạn tham khảo nhé!

\(n_{NaOH}=0,03.0,1=0,003\left(mol\right)\\ n_{HNO_3}=0,01.0,01=0,0001\left(mol\right)\\ NaOH+HNO_3\rightarrow NaNO_3+H_2O\\ Vì:\dfrac{0,0001}{1}< \dfrac{0,003}{1}\\ \Rightarrow NaOHdư\\ n_{NaOH\left(dư\right)}=0,003-0,0001=0,0029\left(mol\right)\\ \left[OH^-\left(dư\right)\right]=\left[NaOH_{dư}\right]=\dfrac{0,0029}{0,01+0,1}=\dfrac{29}{1100}\left(M\right)\\ \Rightarrow pH=14+log\left[\dfrac{29}{1100}\right]\approx12,421\)

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

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

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

Lập tỉ lệ : 

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

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

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

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

\(b.\)

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

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

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

\(n_{H^+}=0,2.\left(0,02+0,01.2\right)=0,008\left(mol\right)\)

\(n_{OH^-}=0,3.2.0,04=0,024\left(mol\right)\)

\(n_{OH^-dư}=0,3.2.0,04=0,016\left(mol\right)\)

\(\Rightarrow\left[OH^-_{dư}\right]=\dfrac{0,016}{0,5}=0,032M\)

\(\Rightarrow\left[H^+\right]=3,125.10^{-13}M\)

\(\Rightarrow pH\approx12,5\)

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 : \(0,01>0,002\Rightarrow H^+dư\)

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

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

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

nNaOH=0,1.0,01=0,001(mol)

nHCl=0,1.0,012=0,0012(mol)

NaOH + HCl\(\rightarrow\)NaCl + H2O

nHCl dư=0,0012-0,001=0,0002(mol)

CMH⁺=\(\frac{0,0002}{0,2}\)= 0,001(M)

pH=-log(0,001)=3