Bài 2: Axit - Bazơ - Muối

Ta có: \(n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,2.1+2.0,2.0,5=0,4\left(mol\right)\)

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

⇒ n_OH^- = n_H⁺ = 0,4 (mol)

Mà: \(n_{OH^-}=n_{NaOH}+2n_{Ba\left(OH\right)_2}=0,3.0,5+2.0,3.a\left(mol\right)\)

⇒ 0,3.0,5 + 2.0,3.a = 0,4 ⇒ a = 5/12 (M)

\(n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,3.\dfrac{5}{12}=0,125\left(mol\right)\)

\(n_{SO_4^{2-}}=n_{H_2SO_4}=0,2.0,5=0,1\left(mol\right)\)

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

\(\Rightarrow n_{BaSO_4}=0,1\left(mol\right)\Rightarrow m=m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

Ta có: \(pH=-log\left[H^+\right]=3\Rightarrow\left[H^+\right]=10^{-3}\left(M\right)\)

\(\Rightarrow n_{HCl}=n_{H^+}=10^{-3}.0,01=10^{-5}\left(mol\right)\)

\(pH=4=-log\left[H^+\right]\) ⇒ [H⁺] _{sau pha} = 10^-4 (M) = [HCl] _{sau pha}

⇒ V_{HCl sau pha} = \(\dfrac{10^{-5}}{10^{-4}}=0,1\left(l\right)=100\left(ml\right)\)

→ Cần thêm 90 ml nước vào dd chứa 10 ml HCl pH = 3 để được dd có pH = 4

a) \(n_{NO}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: \(3Mg+8HNO_3\rightarrow3Mg\left(NO_3\right)_2+2NO+4H_2O\)

            0,15<--------------------------------------0,1

             \(Fe_2O_3+6HNO_3\rightarrow2Fe\left(NO_3\right)_3+3H_2O\)

b) \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,15.24}{20}.100\%=18\%\\\%m_{Fe_2O_3}=100\%-18\%=82\%\end{matrix}\right.\)

Câu 5: \(n_{NaOH}=\dfrac{0,8}{40}=0,02\left(mol\right);V_{dd}=200\left(ml\right)=0,2\left(l\right)\)

`=>` \(C_{M\left(NaOH\right)}=\dfrac{0,02}{0,2}=0,1M\Rightarrow\left[OH^-\right]=0,1=10^{-1}\)

`=>` \(pOH=1\Rightarrow pH=14-1=13\)

Câu 6:

Đặt `V_{ddNaOH} = x (l)`

`=>` \(\left\{{}\begin{matrix}n_{HCl}=0,094x\left(mol\right)\\n_{NaOH}=0,2.0,2=0,04\left(mol\right)\end{matrix}\right.\)

Vì dd sau phản ứng có `pH = 2 < 7 =>` axit dư

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

              0,04---->0,04

`=>` \(n_{H^+\left(dư\right)}=n_{HCl\left(dư\right)}=0,094x-0,04\left(mol\right)\)

Ta có: \(pH=2\Rightarrow\left[H^+\right]=0,01\)

`=>` \(\left[H^+_{dư}\right]=\dfrac{0,094x-0,04}{0,2+x}=0,01\Leftrightarrow x=0,5\left(l\right)\)

`=> V = 500 (ml)`

Câu 8:

Ta có: \(\left\{{}\begin{matrix}n_{Ba\left(OH\right)_2}=0,005.0,02=0,0001\left(mol\right)\\n_{H_2SO_4}=0,08.0,002=0,00016\left(mol\right)\end{matrix}\right.\)

PTHH: \(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2H_2O\)

            0,0001----->0,0001

`=>` \(n_{H^+\left(dư\right)}=2n_{H_2SO_4\left(dư\right)}=\left(0,00016-0,0001\right).2=0,00012\left(mol\right)\)

`=>` \(\left[H^+\right]=\dfrac{0,00012}{0,02+0,08}=0,0012\left(mol\right)\)

`=>` \(pH=-log\left(0,0012\right)=2,921\)

\(n_{H_3PO_3}=0,0115\left(mol\right)\)

\(H_3PO_3+2NaOH\rightarrow Na_2HPO_3+2H_2O\)

0,0115   →   0,023

\(\Rightarrow V_{NaOH}=\dfrac{0,023}{2,3}=0,01\left(l\right)=10\left(ml\right)\)

\(n_{H_3PO_3}=\dfrac{0.943}{82}=0.0115\left(mol\right)\)

\(H_3PO_3\) là axit 2 nấc => phản ứng với dung dịch NaOH theo tỉ lệ 1 : 2 

\(\Rightarrow n_{NaOH}=0.0115\cdot2=0.023\left(mol\right)\)

\(V_{dd_{NaOH}}=\dfrac{0.023}{2.3}=0.01\left(l\right)\)

\(NaH_2PO_3\) là muối axit, \(Na_2HPO_3\) là muối trung hoà.

Gọi \(\left\{{}\begin{matrix}C_{M\left(H_2SO_4\right)}=aM\\C_{M\left(NaOH\right)}=bM\end{matrix}\right.\)

TN₁: Vì dd sau phản ứng cần dùng HCl để trung hoà `=>` NaOH dư

Ta có: \(\left\{{}\begin{matrix}n_{H_2SO_4}=0,01a\left(mol\right)\\n_{NaOH}=0,01b\left(mol\right)\\n_{HCl\left(thêm\right)}=0,2.0,05=0,01\left(mol\right)\end{matrix}\right.\)

PTHH: \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\) (1)

            \(HCl+NaOH\rightarrow NaCl+H_2O\) (2)

Theo PT: \(n_{NaOH}=2n_{H_2SO_4}+n_{HCl}\)

`=>` \(0,01b-0,02a=0,01\left(1\right)\)

TN₂: Vì dd sau cần thêm NaOH để trung hoà => axit dư

Ta có: \(\left\{{}\begin{matrix}n_{H_2SO_4}=0,075a\left(mol\right)\\n_{NaOH}=0,025b+0,025\left(mol\right)\end{matrix}\right.\)

Theo PT: \(n_{NaOH}=2n_{H_2SO_4}\)

`=>` \(0,15a-0,025b=0,025\left(2\right)\)

Từ \(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=0,5M\\b=2M\end{matrix}\right.\)

Bài 6 : 

a) $n_{Al_2(SO_4)_3} = 0,05.1 = 0,05(mol);  n_{NaOH} = 0,2.1,8 = 0,36(mol)$

\(Al_2\left(SO_4\right)_3+6NaOH\text{→}2Al\left(OH\right)_3+3Na_2SO_4\)

0,05                        0,3               0,1                                    (mol)

\(Al\left(OH\right)_3+NaOH\text{→}NaAlO_2+2H_2O\)

0,06                   0,06                                          (mol)

$\Rightarrow n_{Al(OH)_3} = 0,1 - 0,06 = 0,04(mol)$
$\Rightarrow m_{Al(OH)_3} = 0,04.78 = 3,12(gam)$

b) $n_{Al_2(SO_4)_3} = 0,05(mol);  n_{Ba(OH)_2} = 0,1(mol)$

\(Al_2\left(SO_4\right)_3+3Ba\left(OH\right)_2\text{→}2Al\left(OH\right)_3+3BaSO_4\)

\(\dfrac{0,1}{3}\)                           0,1              \(\dfrac{0,2}{3}\)               0,1             (mol)

$m_{kết\ tủa} = m_{Al(OH)_3} + m_{BaSO_4} = \dfrac{0,2}{3}.78 + 0,1.233 = 28,5(gam)$

Bài 4 : 

a)

$n_{NaOH} = 0,1.0,2 = 0,02(mol) ; n_{KOH} = 0,1.0,5 = 0,05(mol)$
$V_{dd} = 0,1 + 0,1 = 0,2(lít)$
$C_{M_{NaOH}} = \dfrac{0,02}{0,2} = 0,1M$

$C_{M_{KOH}} = \dfrac{0,05}{0,2} = 0,25M$

Suy ra : 

$[Na^+] = 0,1M ; [K^+] = 0,25M ; [OH^-] = 0,1 + 0,25 = 0,35M$

b)$H^+ + OH^- \to H_2O$

$n_{H^+} = n_{OH^-} = 0,35.0,2 = 0,07(mol)$
$n_{H_2SO_4} = 0,07 : 2 = 0,035(mol)$
$V_{dd\ H_2SO_4} = \dfrac{0,035}{1} = 0,035(lít)$