\(\left\{{}\begin{matrix}n_{NaOH}=0,3.2=0,6\left(mol\right)\\n_{Fe_2\left(SO_4\right)_3}=0,2.0,5=0,1\left(mol\right)\\n_{H_2SO_4}=0,2.0,5=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}n_{Na^+}=0,6\left(mol\right)\\n_{Fe^{3+}}=0,1.2=0,2\left(mol\right)\\n_{H^+}=0,1.2=0,2\left(mol\right)\end{matrix}\right.\\\left\{{}\begin{matrix}n_{SO_4^{2-}}=0,1.3+0,1=0,4\left(mol\right)\\n_{OH^-}=0,6\left(mol\right)\end{matrix}\right.\end{matrix}\right.\)

PT ion rút gọn:

\(H^++OH^-\rightarrow H_2O\)

0,2-->0,2

\(Fe^{3+}+3OH^-\rightarrow Fe\left(OH\right)_3\downarrow\)

\(\dfrac{2}{15}\)<----0,4--------->\(\dfrac{2}{15}\)

\(\Rightarrow m=\dfrac{2}{15}.107=\dfrac{214}{15}\left(g\right)\)

dd sau phản ứng có: \(\left\{{}\begin{matrix}n_{Na^+}=0,6\left(mol\right)\\n_{Fe^{3+}\left(d\text{ư}\right)}=0,2-\dfrac{2}{15}=\dfrac{1}{15}\left(mol\right)\\n_{SO_4^{2-}}=0,4\left(mol\right)\end{matrix}\right.\)

\(V_{\text{dd}}=0,3+0,2=0,5\left(l\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C_{Na^+}=\dfrac{0,6}{0,5}=1,2M\\C_{Fe^{3+}}=\dfrac{\dfrac{1}{15}}{0,5}=\dfrac{2}{15}M\\C_{SO_4^{2-}}=\dfrac{0,4}{0,5}=0,8M\end{matrix}\right.\)

\(C_{M_{HCl}}=a\left(M\right),C_{M_{H_2SO_4}}=b\left(M\right)\)

\(n_{HCl}=a\left(mol\right),n_{H_2SO_4}=b\left(mol\right)\)

\(n_{NaOH}=0.4\cdot0.5=0.2\left(mol\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

\(a..........a.........a\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

\(2b............b..........b\)

\(n_{NaOH}=a+2b=0.2\left(mol\right)\left(1\right)\)

\(m_{muối}=58.5a+142b=12.95\left(g\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.1,b=0.05\)

\(\left[H^+\right]=0.1+0.05\cdot2=0.2\left(M\right)\)

\(\left[Cl^-\right]=0.1\left(M\right)\)

\(\left[SO_4^{2-}\right]=0.05\left(M\right)\)

\(b.\)

\(pH=-log\left(0.2\right)=0.7\)

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)

⇒ n_{OH^- (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: \(n_{H_2SO_4}=0,3.0,5=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{H^+}=2n_{H_2SO_4}=0,3\left(mol\right)\\n_{SO_4^{2-}}=n_{H_2SO_4}=0,15\left(mol\right)\end{matrix}\right.\)

\(n_{K^+}=n_{OH^-}=n_{KOH}=0,2.1=0,2\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

0,3 ___ 0,2 __________ (mol)

\(\Rightarrow n_{H^+\left(dư\right)}=0,1\left(mol\right)\)

⇒ Dung dịch A gồm: H⁺; SO₄^2- 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!

a, \(NaHCO_3+NaOH\rightarrow Na_2CO_3+H_2O\)

\(HCO_3^-+OH^-\rightarrow CO_3^{2-}+H_2O\)

b, \(\left[Na^+\right]=\dfrac{0,2.1,5+0,12.1,6}{0,2+0,12}=1,5376M\)

\(\left[CO_3^{2-}\right]=\dfrac{0,2.1,5}{0,2+0,12}=0,9375M\)

\(n_{H^+}=0,3\left(mol\right)\)

\(n_{OH^-}=0,192\left(mol\right)\)

\(\Rightarrow n_{H^+dư}=0,108\left(mol\right)\)

\(\Rightarrow\left[H^+\right]=\dfrac{0,108}{0,2+0,12}=0,3375M\)

\(\Rightarrow pH\approx0,47\)

B4:

nNaOH = 0,3 . 1,5 + 0,4 . 2,5 = 1,45 (mol)

VddNaOH = 0,3 + 0,4 = 0,7 (l)

CMddNaOH = 1,45/0,7 = 2,07M

B5:

nHCl (sau khi pha) = 0,5 . 2 = 1 (mol)

Gọi VHCl (0,2) = x (l); VHCl (0,8) = y (l)

x + y = 2 (1)

nHCl (0,2) = 0,2x (mol)

nHCl (0,8) = 0,8y (mol)

=> 0,2x + 0,8y = 1 (2)

(1)(2) => x = y = 1 (l)

200ml = 0,2l

\(n_{Ba\left(OH\right)2}=0,5.0,2=0,1\left(mol\right)\)

Pt : \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O|\)

              1              2             1            2

             0,1           0,2          0,1

a) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

\(V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)

b) \(n_{BaCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{BaCl2}=0,1.208=20,8\left(g\right)\)

c) \(V_{ddspu}=0,2+0,2=0,4\left(l\right)\)

\(C_{M_{BaCl2}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\)

 Chúc bạn học tốt

PTHH: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Ta có: \(n_{Ba\left(OH\right)_2}=0,2\cdot0,5=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{BaCl_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\\m_{BaCl_2}=0,1\cdot208=20,8\left(g\right)\\C_{M_{BaCl_2}}=\dfrac{0,1}{0,2+0,2}=0,25\left(M\right)\end{matrix}\right.\)