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a, \(NaHCO_3+NaOH\rightarrow Na_2CO_3+H_2O\)
\(HCO_3^-+OH^-\rightarrow CO_3^{2-}+H_2O\)
b, \(\left[Na^+\right]=\dfrac{0,2.1,5+0,12.1,6}{0,2+0,12}=1,5376M\)
\(\left[CO_3^{2-}\right]=\dfrac{0,2.1,5}{0,2+0,12}=0,9375M\)
\(n_{H^+}=0,3\left(mol\right)\)
\(n_{OH^-}=0,192\left(mol\right)\)
\(\Rightarrow n_{H^+dư}=0,108\left(mol\right)\)
\(\Rightarrow\left[H^+\right]=\dfrac{0,108}{0,2+0,12}=0,3375M\)
\(\Rightarrow pH\approx0,47\)
\(n_{OH^-}=n_{NaOH}=0,3.1,5=0,45\left(mol\right)\\ n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,2x+0,5.0,2.2=0,2x+0,2\left(mol\right)\)
PT ion rút gọn: \(H^++OH^-\rightarrow H_2O\)
0,45<---0,45
\(\Rightarrow0,2x+0,2=0,45\Leftrightarrow x=1,25M\)
Ta có: \(V_{dd}=0,3+0,2=0,5\left(l\right)\) và \(\left\{{}\begin{matrix}n_{Na^+}=0,45\left(mol\right)\\n_{Cl^-}=0,2.1,25=0,25\left(mol\right)\\n_{SO_4^{2-}}=0,2.0,5=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C_{Na^+}=\dfrac{0,45}{0,5}=0,9M\\C_{Cl^-}=\dfrac{0,25}{0,5}=0,5M\\C_{SO_4^{2-}}=\dfrac{0,1}{0,5}=0,2M\end{matrix}\right.\)
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
=> \(V_{H_2SO_4}=\dfrac{0,05}{1}=0,05\left(l\right)\)
\(n_{NaCl}=0,1.0,2=0,02\left(mol\right)\)
\(m_{Na_2CO_3}=0,1.0,3=0,03\left(mol\right)\)
\(\left\{{}\begin{matrix}n_{Na^+}=0,02+0,03.2=0,08\left(mol\right)\\n_{Cl^-}=0,02\left(mol\right)\\n_{CO_3^{2-}}=0,03\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}C_{M\left(Na^+\right)}=\dfrac{0,08}{0,2+0,3}=0,16M\\C_{M\left(Cl^-\right)}=\dfrac{0,02}{0,2+0,3}=0,04M\\C_{M\left(CO_3^{2-}\right)}=\dfrac{0,03}{0,2+0,3}=0,06M\end{matrix}\right.\)
\(n_{NaOH}=0.25\cdot2=0.5\left(mol\right)\)
\(n_{H_2SO_4}=0.25\cdot1=0.25\left(mol\right)\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)
\(0.5..............0.25................0.25\)
\(\left[Na^+\right]=\dfrac{0.25\cdot2}{0.25+0.25}=1\left(M\right)\)
\(\left[SO_4^{2-}\right]=\dfrac{0.25}{0.25+0.25}=0.5\left(M\right)\)
\(n_{KOH}=0.1\cdot1=0.1\left(mol\right)\)
\(n_{H_2SO_4}=0.3\cdot0.5=0.15\left(mol\right)\)
\(2KOH+H_2SO_4\rightarrow K_2SO_4+H_2O\)
\(0.1..........0.05...............0.05\)
Dung dịch D : 0.05 (mol) K2SO4 , 0.1 (mol) H2SO4
\(\left[K^+\right]=\dfrac{0.05\cdot2}{0.1+0.3}=0.25\left(M\right)\)
\(\left[H^+\right]=\dfrac{0.1\cdot2}{0.1+0.3}=0.5\left(M\right)\)
\(\left[SO_4^{2-}\right]=\dfrac{0.05+0.1}{0.1+0.3}=0.375\left(M\right)\)
\(2NaOH+H_2SO_4\rightarrow K_2SO_4+H_2O\)
\(0.2..................0.1\)
\(V_{dd_{NaOH}}=\dfrac{0.2}{1}=0.2\left(l\right)\)
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!
câu 1 chia 0,6 và câu 2 chia 0,5 là những số đó từ đâu bạn ?