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a) \([OH^-]=\left[KOH\right]=1,5M\)
b) Để trung hòa dung dịch A: \(n_{\left[OH^-\right]}=n_{\left[H^+\right]}\)
\(\Rightarrow0,15=0,5\cdot V_{HCl}\Rightarrow V_{HCl}=0,3l=300ml\)
\(n_{KOH}=0.1\cdot1=0.1\left(mol\right)\)
\(n_{HCl}=0.1\cdot0.5=0.05\left(mol\right)\)
\(KOH+HCl\rightarrow KCl+H_2O\)
\(0.05.......0.05.......0.05\)
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)
$KOH + HCl \to KCl + H_2O$
$n_{KOH} = 0,3(mol) < n_{HCl} = 1,05(mol)$ nên HCl dư
$n_{HCl\ dư} = 1,05 -0 ,3 = 0,75(mol)$
$n_{KCl} = n_{KOH} = 0,3(mol)$
$V_{dd} = 0,3+ 0,7 = 1(lít)$
Suy ra :
$[K^+] = \dfrac{0,3}{1} = 0,3M$
$[Cl^-] = \dfrac{0,75 + 0,3}{1} = 1,05M$
$[H^+] = \dfrac{0,75}{1} = 0,75M$
b)
$Ba(OH)_2 + 2HCl \to BaCl_2 + 2H_2O$
$n_{Ba(OH)_2} = \dfrac{1}{2}n_{HCl} = 0,375(mol)$
$V_{Ba(OH)_2} = \dfrac{0,375}{1,5} = 0,25(lít)$
\(n_{KOH}=0.3\cdot1=0.3\left(mol\right)\)
\(n_{HCl}=0.7\cdot1.5=1.05\left(mol\right)\)
\(KOH+HCl\rightarrow KCl+H_2O\)
\(0.3...........0.3..........0.3\)
Dung dịch D gồm : 0.3 (mol) KCl , 0.75 (mol) HCl dư
\(\left[K^+\right]=\dfrac{0.3}{0.3+0.7}=0.3\left(M\right)\)
\(\left[Cl^-\right]=\dfrac{0.3+0.75}{0.3+0.7}=1.05\left(M\right)\)
\(\left[H^+\right]=\dfrac{0.75}{0.3+0.7}=0.75\left(M\right)\)
\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
\(0.375..................0.75\)
\(V_{dd_{Ba\left(OH\right)_2}}=\dfrac{0.375}{1.5}=0.25\left(l\right)\)
[K+]=[Cl-]=0,25M
[KOH dư]=0,25M
b) 2KCl + H2SO4 ----------->K2SO4 + 2HCl
0,05(mol)---->0,025(mol)
=>vH2SO4=\(\frac{0,025}{1}\)=0,025(lít)
c)pH=-log(0,25)=0,602
(câu c mình không chắc chắn lắm nha bạn!!!)
Cho mình hỏi s [K+]=[Cl-]=0,25M đc z. Mình chưa hiểu lắm
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?