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nH+(1)=0,2(mol)
nH+(2)=0,6(mol)
tổng số mol sau khi trộn là 0,2+0,6=0,8 (mol)
tổng thể tích sau khi trộn là 0,5 (lít)
nồng độ dd sau khi trộn là CM=0,8/0,5=1,6(M)
Bài 1:
Ta có: \(n_{OH^-}=n_{Na^+}=n_{NaOH}=0,2.0,4=0,08\left(mol\right)\)
\(n_{H^+}=n_{Cl^-}=n_{HCl}=0,4.0,3=0,12\left(mol\right)\)
PT ion: \(OH^-+H^+\rightarrow H_2O\)
_____0,08_____0,12 (mol)
⇒ nOH- (dư) = 0,04 (mol)
\(\Rightarrow\left\{{}\begin{matrix}\left[Na^+\right]=\frac{0,08}{0,6}\approx0,133M\\\left[Cl^-\right]=\frac{0,12}{0,6}=0,2M\\\left[OH^-\right]=\frac{0,04}{0,6}\approx0,066M\end{matrix}\right.\)
Câu 2:
Ta có: \(\Sigma n_{K^+}=n_{KCl}+2n_{K_2SO_4}=0,2.1,5+0,3.2.2=1,5\left(mol\right)\)
\(n_{Cl^-}=n_{KCl}=0,2.1,5=0,3\left(mol\right)\)
\(n_{SO_4^{2-}}=0,3.2=0,6\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\left[K^+\right]=\frac{1,5}{0,5}=3M\\\left[Cl^-\right]=\frac{0,3}{0,5}=0,6M\\\left[SO_4^{2-}\right]=\frac{0,6}{0,5}=1,2M\end{matrix}\right.\)
Bạn tham khảo nhé!
a) Ta có: \(\left\{{}\begin{matrix}n_{CuCl_2}=0,3\left(mol\right)\\n_{BaCl_2}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left[Cu\right]=\dfrac{0,3}{0,5}=0,6\left(M\right)\\ \Rightarrow\left[Ba\right]=\dfrac{0,1}{0,5}=0,2\left(M\right)\\ \Rightarrow\left[Cl\right]=\dfrac{0,3.2+0,1.2}{0,5}=1,6\left(M\right)\)
Câu 1:
PT ion: \(H^++OH^-\rightarrow H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{OH^-}=0,6\cdot0,4+0,6\cdot0,3\cdot2=0,6\left(mol\right)\\n_{H^+}=0,2\cdot2,6=0,52\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\) H+ hết, OH- còn dư \(\Rightarrow n_{OH^-\left(dư\right)}=0,08\left(mol\right)\)
\(\Rightarrow\left[OH^-\right]=\dfrac{0,08}{0,6+0,2}=0,1\left(M\right)\) \(\Rightarrow pH=14+log\left(0,1\right)=13\)
Bài 2:
PT ion: \(H^++OH^-\rightarrow H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{OH^-}=0,3\cdot1,6=0,48\left(mol\right)\\n_{H^+}=0,2\cdot1\cdot2+0,2\cdot2=0,8\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\) OH- hết, H+ còn dư \(\Rightarrow n_{H^+\left(dư\right)}=0,32\left(mol\right)\)
\(\Rightarrow\left[H^+\right]=\dfrac{0,32}{0,2+0,3}=0,64\left(M\right)\) \(\Rightarrow pH=-log\left(0,64\right)\approx0,19\)
\(n_{KOH}=0.1\cdot1=0.1\left(mol\right)\)
\(n_{HCl}=0.1\cdot0.5=0.05\left(mol\right)\)
\(KOH+HCl\rightarrow KCl+H_2O\)
\(0.05.......0.05.......0.05\)
Dung dịch D : 0.05 (mol) KOH , 0.05 (mol) KCl
\(\left[K^+\right]=\dfrac{0.05+0.05}{0.1+0.1}=0.5\left(M\right)\)
\(\left[Cl^-\right]=\dfrac{0.05}{0.2}=0.25\left(M\right)\)
\(\left[OH^-\right]=\dfrac{0.05}{0.2}=0.25\left(M\right)\)
\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
\(0.05.........0.025\)
\(V_{dd_{H_2SO_4}}=\dfrac{0.025}{1}=0.025\left(l\right)\)
\(n_{KOH}=0.1\cdot1=0.1\left(mol\right)\)
\(n_{H_2SO_4}=0.3\cdot0.5=0.15\left(mol\right)\)
\(2KOH+H_2SO_4\rightarrow K_2SO_4+H_2O\)
\(0.1..........0.05...............0.05\)
Dung dịch D : 0.05 (mol) K2SO4 , 0.1 (mol) H2SO4
\(\left[K^+\right]=\dfrac{0.05\cdot2}{0.1+0.3}=0.25\left(M\right)\)
\(\left[H^+\right]=\dfrac{0.1\cdot2}{0.1+0.3}=0.5\left(M\right)\)
\(\left[SO_4^{2-}\right]=\dfrac{0.05+0.1}{0.1+0.3}=0.375\left(M\right)\)
\(2NaOH+H_2SO_4\rightarrow K_2SO_4+H_2O\)
\(0.2..................0.1\)
\(V_{dd_{NaOH}}=\dfrac{0.2}{1}=0.2\left(l\right)\)
a, \(K_2SO_4\rightarrow2K^++SO_4^{2-}\)
___0,5_______1______0,5_ (mol)
\(\Rightarrow\left\{{}\begin{matrix}\left[K^+\right]=\frac{1}{2}=0,5M\\\left[SO_4^{2-}\right]=\frac{0,5}{2}=0,25M\end{matrix}\right.\)
b, Ta có: \(n_{OH^-}=n_{K^+}=n_{KOH}=0,2.1=0,2\left(mol\right)\)
\(n_{H^+}=n_{Cl^-}=0,1.1=0,1\left(mol\right)\)
PT ion: \(OH^-+H^+\rightarrow H_2O\)
______0,2_____0,1_________ (mol)
⇒ OH- dư. ⇒ nOH- (dư) = 0,1 (mol)
Dd X gồm: K+; Cl- và OH-(dư).
\(\Rightarrow\left\{{}\begin{matrix}\left[K^+\right]=\frac{0,2}{0,3}=\frac{2}{3}M\\\left[Cl^-\right]=\frac{0,1}{0,3}=\frac{1}{3}M\\\left[OH^-\right]_{\left(dư\right)}=\frac{0,1}{0,3}=\frac{1}{3}M\end{matrix}\right.\)
c, Ta có: \(\left\{{}\begin{matrix}n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,0005.0,5=0,00025\left(mol\right)\\n_{OH^-}=2n_{Ba\left(OH\right)_2}=2.0,0005.0,5=0,0005\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\Sigma n_{H^+}=n_{HNO_3}+n_{HCl}=1.0,1+1.0,05=0,15\left(mol\right)\\n_{NO_3^-}=n_{HNO_3}=1.0,1=0,1\left(mol\right)\\n_{Cl^-}=n_{HCl}=1.0,05=0,05\left(mol\right)\end{matrix}\right.\)
PT ion: \(OH^-+H^+\rightarrow H_2O\)
____0,0005____0,15_________ (mol)
⇒ H+ dư. ⇒ nH+ (dư) = 0,1495 (mol)
Dd D gồm: Ba2+; NO3-; Cl- và H+(dư)
\(\Rightarrow\left\{{}\begin{matrix}\left[Ba^{2+}\right]=\frac{0,00025}{1,0005}\approx2,5.10^{-4}M\\\left[NO_3^-\right]=\frac{0,1}{1,0005}\approx0,09M\\\left[Cl^-\right]=\frac{0,05}{1,0005}\approx0,049M\\\left[H^+\right]_{\left(dư\right)}=\frac{0,1495}{1,0005}\approx0,15M\end{matrix}\right.\)
Bạn tham khảo nhé!
Mà phần c số lẻ quá, không biết đề là 0,5 ml hay 0,5 lít bạn nhỉ?
a) \(n_{NaOH}=0,2.1=0,2\left(mol\right)\); \(n_{HNO_3}=0,2.0,5=0,1\left(mol\right)\)
\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)
0,2.............0,1
Lập tỉ lệ : \(\dfrac{0,2}{1}>\dfrac{0,1}{1}\) => Sau phản ứng NaOH dư
Dung dịch D gồm NaNO3 và NaOH dư
\(n_{NaNO_3}=n_{HNO_3}=0,1\left(mol\right)\)
\(n_{NaOH\left(pứ\right)}=n_{HNO_3}=0,1\left(mol\right)\)
\(n_{NaOH\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)
Ion trong dung dịch D : Na+ , NO3-, OH-
\(\left[Na^+\right]=\dfrac{0,1+0,1}{0,2}=1M\)
\(\left[NO_3^-\right]=\dfrac{0,1}{0,2}=0,5M\)
\(\left[OH^-\right]=\dfrac{0,1}{0,2}=0,5M\)
b)Trong dung dịch D chỉ có NaOH dư phản ứng
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)
0,1................0,05
=> \(V_{H_2SO_4}=\dfrac{0,05}{1}=0,05\left(l\right)\)
Đáp án C
CM HCl= (0,2+0,3.2)/0,5= 1,6M