Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(\left[Ca^{2+}\right]=\dfrac{0,15.0,5}{0,15+0,05}=0,375M\)
\(\left[Na^+\right]=\dfrac{0,05.2}{0,15+0,05}=0,5M\)
\(\left[Cl^-\right]=\dfrac{0,15.2.0,5+0,05.2}{0,15+0,05}=1,25M\)
Mg từ đâu ra vậy bạn.
\(a)\)
\(n_{CuSO_4}=0,5\left(mol\right)\)
\(CuSO_4\left(0,5\right)--->Cu^{2+}\left(0,5\right)+SO_4^{2-}\left(0,5\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\left[Cu^{2+}\right]=\dfrac{0,5}{0,5}=1M\\\left[SO_4^{2-}\right]=\dfrac{0,5}{0,5}=1M\end{matrix}\right.\)
\(b)\)
\(Cu^{2+}\left(0,5\right)+2OH^-\left(1\right)--->Cu\left(OH\right)_2\downarrow\)
\(n_{OH^-}=0,1\left(mol\right)\)
\(\Rightarrow V_{ddKOH}=\dfrac{1}{0,5}=2\left(l\right)\)
\(c)\)
\(Ba^{2+}\left(0,5\right)+SO_4^{2-}\left(0,5\right)--->BaSO_4\downarrow\)
\(n_{Ba^{2+}}=0,5\left(mol\right)\)
\(\Rightarrow V_{ddBaCl_2}=\dfrac{0,5}{0,25}=2\left(l\right)\)
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:
\(n_{H^+}=n_{OH^-}\)
\(\Leftrightarrow0,01.V_{ddHCl}=\left(0,02.2+0,01\right).0,2\)
\(\Leftrightarrow V_{ddHCl}=1\left(l\right)\)
nMgSO4 = 0.2
CM MgSO4 = [Mg2+]=[SO42-]= 0.25M
Mg2 + 2OH- = Mg(OH)2
0.2-----0.4
=> V KOH = 0.8l
nBa(OH)2 = 0.2
=> CM Ba(OH)2 = 0.4 M
[Ba2+] = 0.4M
[OH-]= 2*0.4=0.8 M
H+ + OH - = H2O
0.4----0.4
=> nH2SO4 = 0.2 => V =1
Cảm ơn bạn