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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) ⇒ nH+ dư = 0,05 (mol)
\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)
0,015___0,015______0,015 (mol) ⇒ nSO42- dư = 0,015 (mol)
⇒ m = mBaSO4 = 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, \(n_{H^+}=0,025.0,2=0,005\left(mol\right)\)
\(n_{OH^-}=0,01.2.0,3=0,006\left(mol\right)\)
\(\Rightarrow n_{OH^-dư}=0,001\left(mol\right)\)
\(\Rightarrow\left[OH^-\right]_{dư}=\dfrac{0,001}{1}=10^{-3}\)
\(\Rightarrow\left[H^+\right]=10^{-11}\)
\(\Rightarrow pH=11\)
b, \(n_{Fe^{2+}}=n_{SO_4^{2-}}=0,02.0,1=0,002\left(mol\right)\)
\(n_{Ba^{2+}}=0,01.0,3=0,003\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{BaSO_4\downarrow}=n_{SO_4^{2-}}=0,002\left(mol\right)\\n_{Fe\left(OH\right)_2\downarrow}=n_{OH^-dư}=0,001\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{\downarrow}=0,002.233+0,001.90=0,556\left(g\right)\)
Đáp án C
nMgSO4= 0,1 mol; nAl2(SO4)3= 0,1 mol; Vdd= 0,5 lít
CMgSO4=0,2M; CMAl2(SO4)3= 0,2M
MgSO4 → Mg2++ SO42-
0,2M 0,2M
Al2(SO4)3→ 2Al3++ 3SO42-
0,2M 0,6M
[SO42-] = 0,8M
\(n_{FeCl_3}=0.1\cdot0.1=0.01\left(mol\right)\)
\(n_{NaOH}=0.5\cdot0.1=0.05\left(mol\right)\)
\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(1............3\)
\(0.01...........0.05\)
Lập tỉ lệ : \(\dfrac{0.01}{1}< \dfrac{0.05}{3}\Rightarrow NaOHdư\)
Các chất có trong D : \(NaCl:0.03\left(mol\right),NaOH\left(dư\right):0.02\left(mol\right)\)
\(V=0.1+0.5=0.6\left(l\right)\)
\(\left[Na^+\right]=\dfrac{0.03+0.02}{0.06}=\dfrac{1}{12}\left(M\right)\)
\(\left[Cl^-\right]=\dfrac{0.03}{0.06}=0.5\left(M\right)\)
\(\left[OH^-\right]=\dfrac{0.02}{0.6}=\dfrac{1}{30}\left(M\right)\)
\(b.\)
\(m_{Fe\left(OH\right)_3}=0.01\cdot107=1.07\left(g\right)\)
Đáp án D
nFe2(SO4)3= 0,4.0,2= 0,08 mol; nFeCl3= 0,1.0,3=0,03 mol
CM Fe2(SO4)3= 0,08/0,5=0,16M; CM FeCl3= 0,03/0,5= 0,06M
Fe2(SO4)3 → 2Fe3+ + 3SO42-
0,16M 0,32 M
FeCl3 → Fe3+ + 3Cl-
0,06 M 0,06 M
CM Fe3+= 0,32+ 0,06= 0,38M
a, \(\left[Ca^{2+}\right]=\dfrac{0,15.0,5}{0,15+0,05}=0,375M\)
\(\left[Na^+\right]=\dfrac{0,05.2}{0,15+0,05}=0,5M\)
\(\left[Cl^-\right]=\dfrac{0,15.2.0,5+0,05.2}{0,15+0,05}=1,25M\)