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Đáp án B
n C u S O 4 . 5 H 2 O = 50/250=0,2 mol nên C M C u S O 4 = 0,2/0,2=1M=[Cu2+]= [SO42-]
a, \(\left[Ca^{2+}\right]=\dfrac{0,15.0,5}{0,15+0,05}=0,375M\)
\(\left[Na^+\right]=\dfrac{0,05.2}{0,15+0,05}=0,5M\)
\(\left[Cl^-\right]=\dfrac{0,15.2.0,5+0,05.2}{0,15+0,05}=1,25M\)
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:
\(n_{H^+}=n_{OH^-}\)
\(\Leftrightarrow0,01.V_{ddHCl}=\left(0,02.2+0,01\right).0,2\)
\(\Leftrightarrow V_{ddHCl}=1\left(l\right)\)
a)
$n_{CaCO_3} = 0,12(mol) ; n_{HCl} = 0,6(mol)
\(CaCO_3+2HCl\text{→}CaCl_2+CO_2+H_2O\)
Ban đầu 0,12 0,6 (mol)
Phản ứng 0,12 0,24 (mol)
Sau pư 0 0,36 0,12 (mol)
$V = 0,12.22,4 = 2,688(lít)$
b)
$n_{Cl^-} = 0,6(mol) ; n_{H^+} = 0,36(mol)$
$n_{Ca^{2+}} = 0,12(mol)$
$[Cl^-] = \dfrac{0,6}{0,2} = 3M$
$[H^+] = \dfrac{0,36}{0,2} = 1,8M$
$[Ca^{2+}] = \dfrac{0,12}{0,2} = 0,6M$
a,\(n_{CaCO_3}=\dfrac{12}{100}=0,12\left(mol\right);n_{HCl}=0,2.3=0,6\left(mol\right)\)
PTHH: CaCO3 + 2HCl → CaCl2 + CO2 + H2O
Mol: 0,12 0,12
Ta có: \(\dfrac{0,12}{1}< \dfrac{0,6}{2}\)⇒ HCl dư,CaCO3 pứ hết
\(V_{CO_2}=0,12.22,4=2,688\left(l\right)\)
a) Ta có: \(\left\{{}\begin{matrix}\left[Cu^{2+}\right]=C_{M_{Cu\left(NO_3\right)_2}}=0,3\left(M\right)\\\left[NO_3^-\right]=2C_{M_{Cu\left(NO_3\right)_2}}=0,6\left(M\right)\end{matrix}\right.\)
b) Ta có: \(n_{H_2SO_4}=\dfrac{4,9}{98}=0,05\left(mol\right)\) \(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\left[H^+\right]=0,5\left(M\right)\\\left[SO_4^{2-}\right]=0,25\left(M\right)\end{matrix}\right.\)
\(n_{NaOH}=0,02.2=0,04\left(mol\right)\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ a.n_{H_2SO_4}=n_{Na_2SO_4}=\dfrac{0,04}{2}=0,02\left(mol\right)\\ C_{MddH_2SO_4}=\dfrac{0,02}{0,08}=0,25\left(M\right)\\ b.\left[Na^+\right]=\dfrac{0,02.2}{0,02+0,08}=0,4\left(M\right)\\ \left[SO^{2-}_4\right]=\dfrac{0,02}{0,02+0,08}=0,2\left(M\right)\)
a) Ta có: \(n_{NaCl}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\)
\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,1}{0,5}=0,2\left(M\right)=\left[Na^+\right]=\left[Cl^-\right]\)
b) Ta có: \(n_{Ba\left(OH\right)_2}=\dfrac{34,2}{171}=0,2\left(mol\right)\)
\(\Rightarrow C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\) \(\Rightarrow\left\{{}\begin{matrix}\left[Ba^{2+}\right]=0,4\left(M\right)\\\left[OH^-\right]=0,8\left(M\right)\end{matrix}\right.\)
c) Ta có: \(n_{H_2SO_4}=0,025\cdot2=0,05\left(mol\right)\)
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,05}{0,125+0,025}\approx0,33\left(M\right)\) \(\Rightarrow\left\{{}\begin{matrix}\left[H^+\right]=0,66\left(M\right)\\\left[SO_4^{2-}\right]=0,33\left(M\right)\end{matrix}\right.\)
50 gam CuSO4.5H2O có số mol = 0.2 mol
CuSO4 > Cu(2+) + SO4(2-)
0.2----------0.2---------0.2
Cm (Cu2+) = Cm SO4(2-) = 0.2/0.2 = 1M
---------------------------------------...
pH = 2 => [H+] = 0.01M => nH+ = 0.01x0.2 = 0.002 mol
pH = 1 => [H+] = 0.1M => nH+ = 0.1x0.2 = 0.02 mol
=> nHCl cần = nH+ cần = 0.02-0.002 = 0.018 mol
---------------------------------------...
nCH3COOH = 0.01 mol
Cm dd mới pha = 0.01/0.1 = 0.1M
Ka = [CH3COO-][H+]/[CH3COOH] = 4.70 <=> a^2/(0.1-a) = 4.70 <=> a^2 = 0.47 - 4.70a <=> a = ...
Tính đc A rồi ra đc pH=-log[a]
còn cách khác k... mình chưa học đến pH