Bài 4: Phản ứng trao đổi trong dung dịch các chất điện li

\(Cu+4HNO_{3\left(đ\right)}\underrightarrow{t^o}Cu\left(NO_3\right)_2+2NO_2+2H_2O\)

Câu 2:

Gọi \(\left\{{}\begin{matrix}n_{Na_2CO_3}=x\left(mol\right)\\n_{Na_2SO_4}=y\left(mol\right)\end{matrix}\right.\)

PTHH:
\(Na_2CO_3+BaCl_2\rightarrow BaCO_3\downarrow+2NaCl\)

\(Na_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2NaCl\)

\(\xrightarrow[]{BT\text{ }SO_4^{2-},CO_3^{2-}}\left\{{}\begin{matrix}n_{BaCO_3}=x\left(mol\right)\\n_{BaSO_4}=y\left(mol\right)\end{matrix}\right.\)

`=>` \(\left\{{}\begin{matrix}106x+142y=\dfrac{6,63}{1,7}=3,9\\197x+233y=6,63\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,01\\y=0,02\end{matrix}\right.\)

`=>` \(\left\{{}\begin{matrix}C_{M\left(Na_2CO_3\right)}=\dfrac{0,01}{0,25}=0,04M\\C_{M\left(Na_2SO_4\right)}=\dfrac{0,02}{0,25}=0,08M\end{matrix}\right.\)

Câu 3:

Gọi \(\left\{{}\begin{matrix}n_{NaOH}=x\left(mol\right)\\n_{Na_2CO_3}=y\left(mol\right)\\n_{Na_2SO_4}=z\left(mol\right)\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}n_{HCl}=0,2.0,08=0,016\left(mol\right)\\n_{BaCl_2}=\dfrac{4,16}{208}=0,02\left(mol\right)\end{matrix}\right.\)

PTHH:

\(NaOH+HCl\rightarrow NaCl+H_2O\\ Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\\ Na_2CO_3+BaCl_2\rightarrow BaCO_3\downarrow+2NaCl\\ Na_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2NaCl\)

\(\xrightarrow[]{BT\text{ }Na^+,Cl^-,CO_3^{2-},SO_4^{2-}}\left\{{}\begin{matrix}x+2y=0,016\\y+z=0,02\\208y+233z=4,48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,0016\\y=0,0072\\z=0,0128\end{matrix}\right.\)

`=>` \(\left\{{}\begin{matrix}C\%_{NaOH}=\dfrac{0,0016.40}{80}.100\%=0,08\%\\C\%_{Na_2CO_3}=\dfrac{0,0072.106}{80}.100\%=0,954\%\\C\%_{Na_2SO_4}=\dfrac{0,0128.142}{80}.100\%=2,272\%\end{matrix}\right.\)

$n_{HCl} = 0,1.0,5 = 0,05(mol)$
$n_{H_2SO_4} = 0,1.0,5 = 0,05(mol)$
$\Rightarrow n_{H^+} = 0,05 + 0,05.2 = 0,15(mol)$
$n_{KOH} = 0,1.0,6 = 0,06(mol)$
$n_{Ba(OH)_2} = 0,1.0,1 = 0,01(mol)$
$\Rightarrow n_{OH^-} = 0,06 + 0,01.2 = 0,08(mol)$
$H^+ + OH^- \to H_2O$
$n_{H^+\ pư} = n_{OH^-} = 0,08(mol) \Rightarrow n_{H^+\ dư} = 0,07(mol)$
$V_{dd\ sau\ pư} = 0,1 + 0,1 = 0,2(lít)$
$[H^+] = \dfrac{0,07}{0,2} = 0,35M$

$pH = -log(0,35) = 0,456$

$n_{SO_4^{2-}} = n_{H_2SO_4} = 0,05 > n_{Ba^{2+}} = n_{Ba(OH)_2} = 0,01$

nên $SO_4^{2-}$ dư

$n_{BaSO_4} = n_{Ba^{2+}} = 0,01(mol) \Rightarrow m = 0,01.233 = 2,33(gam)$

`a)C_[M_[HCl]]=0,0001(M)=10^[-4](M)`

   `=>[H^+ ]=10^[-4](M)`

  `=>pH=4`

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`b)C_[M_[H_2 SO_4]]=0,0005(M)=5.10^[-4](M)`

    `=>[H^+ ]=2.5.10^[-4]=10^[-3](M)`

  `=>pH=3`

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`c)C_[M_[Ba(OH)_2]]=0,0005(M)=5.10^[-4](M)`

   `=>[OH^- ]=2.5.10^[-4]=10^[-3](M)`

  `=>[H^+ ]=[10^[-14]]/[10^[-3]]=10^[-11](M)`

  `=>pH=11`

Câu 1:

a) \(Pb\left(NO_3\right)_2+CaCl_2\rightarrow PbCl_2\downarrow+Ca\left(NO_3\right)_2\)

\(Pb^{2+}+2Cl^-\rightarrow PbCl_2\downarrow\)

b) \(\left(NH_4\right)_2CO_3+2KOH\rightarrow K_2CO_3+2NH_3\uparrow+H_2O\)

\(NH_4^++OH^-\rightarrow NH_3\uparrow+H_2O\)

Câu 2:

a) Ta có: 

\(\left\{{}\begin{matrix}n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,08.0,25.2=0,04\left(mol\right)\\n_{H^+}=n_{HNO_3}+2n_{H_2SO_4}=0,5.0,02+0,02.0,5.2=0,03\left(mol\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,08.0,25=0,02\left(mol\right)\\n_{SO_4^{2-}}=0,02.0,5=0,01\left(mol\right)\end{matrix}\right.\)

PTHH: \(H^++OH^-\rightarrow H_2O\)

         0,03--->0,03

            \(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\downarrow\)

           0,01<---0,01

`=>` \(ddA:\left\{{}\begin{matrix}n_{OH^-\left(dư\right)}=0,04-0,03=0,01\left(mol\right)\\n_{Ba^{2+}\left(dư\right)}=0,02-0,01=0,01\left(mol\right)\\n_{NO_3^-}=n_{HNO_3}=0,02.0,5=0,01\left(mol\right)\end{matrix}\right.\)

Ta có: \(V_{ddA}=0,08+0,02=0,1\left(l\right)\)

`=>` \(\left[OH^-\right]=\left[Ba^{2+}\right]=\left[NO_3^-\right]=\dfrac{0,01}{0,1}=0,1M\)

b) \(pOH=-\text{log}\left(0,1\right)=1\)

`=>` \(pH=14-1=13\)

$n_{KOH} = 0,03.0,1 = 0,003(mol)$
$n_{Ba(OH)_2} = 0,03.0,025 = 0,00075(mol)$

$\Rightarrow n_{OH^-} = 0,003 + 0,00075.2 = 0,0045(mol)$

$n_{H^+\ dư} = 0,05.10^{-1} = 0,005(mol)$

$OH^- + H^+ \to H_2O$
$n_{H^+\ pư} = n_{OH^-} = 0,0045(mol)$

$\Rightarrow n_{H^+} = 0,0045 + 0,005 = 0,0095(mol)$

$n_{H_2SO_4} = \dfrac{1}{2}n_{H^+} = 0,00475(mol)$

$x = \dfrac{0,00475}{0,02} = 0,2375M$

$Ba^{2+} + SO_4^{2-} \to BaSO_4$
$n_{BaSO_4} = n_{SO_4^{2-}} = n_{Ba(OH)_2} = 0,00075(mol)$
$m = 0,00075.233 = 0,17475(gam)$

$(1) Ca(OH)_2 + Na_2SO_3 \to CaSO_3 + 2NaOH$
$(2) NH_4NO_3 + NaOH \to NaNO_3 + NH_3 + H_2O$
$(3) MgCl_2 + 2KOH \to Mg(OH)_2 + 2KCl$

(1) \(CaCl_2+Na_2SO_3\rightarrow CaSO_3\downarrow+2NaCl\)

(2) \(NH_4F+LiOH\rightarrow LiF+NH_3\uparrow+H_2O\)

(3) \(MgSO_4+2KOH\rightarrow Mg\left(OH\right)_2\downarrow+K_2SO_4\)

$Cu(OH)_2$ là chất điện li yếu nên khi viết phương trình ion phải giữ nguyên