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Ban đầu:\(n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,25.\left(0,08+2.0,01\right)=0,025\left(mol\right)\)
\(n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,25a.2=0,5a\)
\(n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,25a\); \(n_{SO_4^{2-}}=n_{H_2SO_4}=0,0025\left(mol\right)\)
Dung dịch sau phản ứng có pH = 12 => pOH = 2
\(\Rightarrow\left[OH^-\right]=10^{-2}=0,01\left(M\right)\Rightarrow n_{OH^-}=0,01.0,5=0,005\left(mol\right)\)Vì pH = 12 > 7 nên \(H^+\) hết, \(OH^-\) còn.
\(H^++OH^-\rightarrow H_2O\)
0,025-->0,025
=> \(n_{OH^-}\text{còn}=0,5a-0,025=0,005\Rightarrow a=0,06\left(mol\text{/}l\right)\)
Từ đó suy ra được \(n_{Ba^{2+}}=0,25.0,06=0,015\left(mol\right)\)
\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\downarrow\)
0,0025<--0,0025
=> \(m=m_{BaSO_4}=0,0025.233=0,5825\left(gam\right)\)
\(a)n_{Ba\left(OH\right)_2}=0,05\cdot0,2\cdot2=0,02mol\\ pH=1\Rightarrow\left[OH^-\right]=0,1M\Rightarrow n_{HCl}=0,1\cdot0,3=0,03mol\\ n_{Ba\left(OH\right)_2}+n_{HCl}=0,02+0,03=0,05mol\\ \Rightarrow C_M=\dfrac{0,05}{0,5}=0,1M\Rightarrow pH=1\)
nNaOH=0,2a mol
\(C_{M_{NaOH}}=\dfrac{0,2a}{1,8+0,2}=\dfrac{0,2a}{0,2}=aM\)
[OH-]=[NaOH]=aM
pH=14+lg[OH-]\(\rightarrow\)13=14+lga\(\rightarrow\)lga=-1\(\rightarrow\)a=10-1=0,1M
\(a.n_{NaOH}=1.0,15=0,15\left(mol\right)\\ n_{KOH}=0,5.0,1=0,05\left(mol\right)\\ \left[Na^+\right]=\left[NaOH\right]=\dfrac{0,15}{0,15+0,1}=0,6\left(M\right)\\ \left[K^+\right]=\left[KOH\right]=\dfrac{0,05}{0,1+0,15}=0,2\left(M\right)\\ \left[OH^-\right]=0,2+0,6=0,8\left(M\right)\\ b.2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ n_{H_2SO_4}=\dfrac{1}{2}.\left(n_{KOH}+n_{NaOH}\right)=\dfrac{0,15+0,05}{2}=0,1\left(mol\right)\\ a=C_{MddH_2SO_4}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)
\(n_{H^+}=n_{HNO_3}=V\)mol
\(n_{OH^-}=n_{NaOH}=0,5.0,2=0,1\) mol
\(H^++OH^-\rightarrow H_2O\)
0,1<--0,1
\(\Rightarrow n_{H^+}=V=0,1\)lít = 100 ml
\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)
0,1 -----> 0,1 ---------->0,1
\(NaNO_3\rightarrow Na^++NO_3^-\)
\(\Rightarrow\left[Na^+\right]=\left[NO_3^-\right]=\dfrac{0,1}{0,1+0,2}=0,33M\)
\(n_{OH^-}=0,2.1+0,2.1.2=0,6\left(mol\right)\)
\(n_{H^+}=V.2+V.1.2=4V\left(mol\right)\)
H+ + OH- --------> H2O (1)
Vì dung dịch A được trung hòa bởi 200 ml dd HNO3 1M
=>Dung dịch A có OH- dư sau phản ứng
\(n_{H^+}=0,2.1=0,2\left(mol\right)\)
H+ + OH- dư --------> H2O
=> \(n_{OH^-\left(dư\right)}=n_{H^+\left(củaHNO3\right)}=0,2\left(mol\right)\)
=> \(n_{OH^-\left(pứ\right)}=0,6-0,2=0,4\left(mol\right)\)
Từ (1) => \(n_{H^+}=n_{OH^-\left(pứ\right)}=0,4\left(mol\right)\)
=> 4V=0,4
=> V= 0,1 (lít)
a, Ta có: \(n_{H_2SO_4}=0,3.0,5=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{H^+}=2n_{H_2SO_4}=0,3\left(mol\right)\\n_{SO_4^{2-}}=n_{H_2SO_4}=0,15\left(mol\right)\end{matrix}\right.\)
\(n_{K^+}=n_{OH^-}=n_{KOH}=0,2.1=0,2\left(mol\right)\)
\(H^++OH^-\rightarrow H_2O\)
0,3 ___ 0,2 __________ (mol)
\(\Rightarrow n_{H^+\left(dư\right)}=0,1\left(mol\right)\)
⇒ Dung dịch A gồm: H+; SO42- và K+
\(\Rightarrow\left\{{}\begin{matrix}\left[H^+\right]=\frac{0,1}{0,5}=0,2M\\\left[SO_4^{2-}\right]=\frac{0,15}{0,5}=0,3M\\\left[K^+\right]=\frac{0,2}{0,5}=0,4M\end{matrix}\right.\)
b, \(H^++OH^-\rightarrow H_2O\)
__0,1 → 0,1 ___________ (mol)
\(\Rightarrow n_{NaOH}=n_{OH^-}=0,1\left(mol\right)\)
\(\Rightarrow V_{NaOH}=\frac{0,1}{0,5}=0,2\left(l\right)\)
Bạn tham khảo nhé!
Hoang Duong
a, \(H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\)
b, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
\(pH=10^{-13}\Rightarrow\left[OH^-_{dư}\right]=10^{-1}\)
\(\Rightarrow n_{OH^-dư}=10^{-1}.0,5=0,05\left(mol\right)\)
\(n_{H^+}=0,2.0,5=0,1\left(mol\right)\)
\(n_{OH^-}=0,3x\left(mol\right)\)
\(\Rightarrow0,3x=0,05+0,1=0,15\)
\(\Rightarrow x=0,5\)