a, \(n_{Ba\left(OH\right)_2}=0,1.0,1=0,01\left(mol\right)=n_{Ba^{2+}}\)

\(\Rightarrow n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,02\left(mol\right)\)

\(n_{NaOH}=0,1.0,1=0,01\left(mol\right)=n_{Na^+}=n_{OH^-}\)

⇒ Σn_OH^- = 0,02 + 0,01 = 0,03 (mol)

\(n_{H_2SO_4}=0,4.0,0175=0,007\left(mol\right)=n_{SO_4^{2-}}\)

\(\Rightarrow n_{H^+}=2n_{H_2SO_4}=0,014\left(mol\right)\)

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

0,014___0,014 (mol) ⇒ n_OH^- dư = 0,03 - 0,014 = 0,016 (mol)

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

0,007____0,007_____0,007 (mol) ⇒ n_Ba²⁺ dư = 0,01 - 0,007 = 0,003 (mol)

⇒ m = 0,007.233 = 1,631 (g)

\(\left[OH^-\right]=\dfrac{0,016}{0,1+0,4}=0,032\left(M\right)\)

\(\left[Ba^{2+}\right]=\dfrac{0,003}{0,1+0,4}=0,006\left(M\right)\)

\(\left[Na^+\right]=\dfrac{0,01}{0,1+0,4}=0,02\left(M\right)\)

b, pH = 14 - (-log[OH^-]) ≃ 12,505

\(n_{Ba^{2+}}=0,1.0,1=0,01\left(mol\right)\)

\(n_{SO_4^{2-}}=0,4.0,0175=7.10 ^{-3}\left(mol\right)\)

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

\(\Rightarrow m=m_{BaSO_4}=7.10^{-3}.233=1,631\left(g\right)\)

Ta có:

\(n_{H^+}=0,4.0,0175.2=0,014\left(mol\right)\)

\(n_{OH^-}=0,1.0,1.2+0,1.0,1=0,03\left(mol\right)\)

Trong dung dịch X:

\(n_{OH^-}=0,03-0,014=0,016\left(mol\right)\)\(\Rightarrow\left[OH^-\right]=\dfrac{0,016}{0,1+0,4}=0,032\left(M\right)\)

\(n_{Ba^{2+}}=0,01-7.10^{-3}=3.10^{-3}\left(mol\right)\Rightarrow\left[Ba^{2+}\right]=\dfrac{3.10^{-3}}{0,1+0,4}=6.10^{-3}\left(M\right)\)

\(n_{Na^+}=0,1.0,1=0,01\left(mol\right)\Rightarrow\left[Na^+\right]=0,02\)

\(pOH=-lg\left(0,032\right)\approx1,5\Rightarrow pH=14-1,5=12,5\)

a, \(n_{HCl}=0,2.0,1=0,02\left(mol\right)=n_{H^+}=n_{Cl^-}\)

\(n_{H_2SO_4}=0,2.0,15=0,03\left(mol\right)=n_{SO_4^{2-}}\) \(\Rightarrow n_{H^+}=2n_{H_2SO_4}=0,06\left(mol\right)\)

\(\Rightarrow\Sigma n_{H^+}=0,02+0,06=0,08\left(mol\right)\)

\(n_{Ba\left(OH\right)_2}=0,3.0,05=0,015\left(mol\right)=n_{Ba^{2+}}\) 

\(\Rightarrow n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,03\left(mol\right)\)

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

0,03___0,03 (mol) ⇒ n_H⁺ dư = 0,05 (mol)

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

0,015___0,015______0,015 (mol) ⇒ n_SO4^2- dư = 0,015 (mol)

⇒ m = m_BaSO4 = 0,015.233 = 3,495 (g)

\(\left[Cl^-\right]=\dfrac{0,02}{0,2+0,3}=0,04\left(M\right)\)

\(\left[H^+\right]=\dfrac{0,05}{0,2+0,3}=0,1\left(M\right)\)

\(\left[SO_4^{2-}\right]=\dfrac{0,015}{0,2+0,3}=0,03\left(M\right)\)

b, pH = -log[H⁺] = 1

a) Ta có: \(n_{Al\left(NO_3\right)_3}=\dfrac{4,26}{213}=0,02\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al^+}=0,02\left(mol\right)\\n_{NO_3^-}=0,06\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[Al^+\right]=\dfrac{0,02}{0,1}=0,2\left(M\right)\\\left[NO_3^-\right]=\dfrac{0,06}{0,1}=0,6\left(M\right)\end{matrix}\right.\)

b) Ta có: \(\left[Na^+\right]=0,1+0,02\cdot2+0,3=0,304\left(M\right)\)

c) Bạn xem lại đề !!

$NaNO_3 \to Na^+ + NO_3^-$
$Na_2SO_4 \to 2Na^+ + SO_4^{2-}$

$NaCl \to Na^+ + Cl^-$

Ta có : 

$[Na^+] = 0,1 + 0,02.2 + 0,3 = 0,44M$

Chọn C.

n O H - = n H + H 2 O = 2 n H 2 = 0 , 04 n O H - = n H + H C l + 2 n C u O H 2 0 , 04             0 , 02                             ? ⇒ n C u O H 2 = 0 , 01   m o l . m C u O H 2 = 0 , 98   g a m .

Chọn đáp án C.

\(n_{NaCl}=0.1\cdot0.1=0.01\left(mol\right)\)

\(n_{AgNO_3}=0.15\cdot0.1=0.015\left(mol\right)\)

\(NaCl+AgNO_3\rightarrow NaNO_3+AgCl\)

\(0.01..........0.01...............0.01\)

Dung dịch : 0.01 (mol) NaNO₃ , 0.005 (mol) AgNO₃

\(\left[Na^+\right]=\dfrac{0.01}{0.1+0.15}=0.04\left(M\right)\)

\(\left[Ag^+\right]=\dfrac{0.005}{0.25}=0.02\left(M\right)\)

\(\left[NO_3^-\right]=\dfrac{0.01+0.005}{0.25}=0.06\left(M\right)\)

\(n_{NaCl}=0,1.0,2=0,02\left(mol\right)\)

\(m_{Na_2CO_3}=0,1.0,3=0,03\left(mol\right)\)

\(\left\{{}\begin{matrix}n_{Na^+}=0,02+0,03.2=0,08\left(mol\right)\\n_{Cl^-}=0,02\left(mol\right)\\n_{CO_3^{2-}}=0,03\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}C_{M\left(Na^+\right)}=\dfrac{0,08}{0,2+0,3}=0,16M\\C_{M\left(Cl^-\right)}=\dfrac{0,02}{0,2+0,3}=0,04M\\C_{M\left(CO_3^{2-}\right)}=\dfrac{0,03}{0,2+0,3}=0,06M\end{matrix}\right.\)

Chọn đáp án A

pH ≈ 5 có nghĩa môi trường là axit do đó chỉ A hoặc C thỏa mãn .Tuy nhiên BaSO₄ kết tủa