Bài 3: Sự điện li của H2O - pH - Chất chỉ thị axit-bazo

Vì dd sau phản ứng có pH = 1 => dd sau phản ứng chứa axit dư

Ta có: \(\left\{{}\begin{matrix}n_{HNO_3}=0,6.1=0,6\left(mol\right)\\n_{KOH}=0,4a\left(mol\right)\end{matrix}\right.\)

Lại có: \(pH=1\Rightarrow\left[H^+_{dư}\right]=10^{-1}=0,1M\)

`=>` \(n_{HNO_3\left(dư\right)}=n_{H^+\left(dư\right)}=0,1.1=0,1\left(mol\right)\)

PTHH: \(KOH+HNO_3\rightarrow KNO_3+H_2O\)

             0,4a---->0,4a

`=>` \(0,6-0,4a=0,1\Leftrightarrow a=1,25M\)

$\rm{pH=1\to Moi\,truong\,axit\,hay\,axit \,du\\\to \Sigma [H^+]-\Sigma [OH^-]=1\times 10^{-1}=0,1(mol)\\\to \Sigma n_{H^+}=0,6\times 1 =0,6(mol) \\\to \Sigma n_{OH^-}=0,4\times a=0,4a\\\to \dfrac{0,6-0,4a}{1}=0,1\\\to a =1,25}$

\(n_{OH-}=0.01V\)

\(n_{H+}=0.03V\)

=>H+ dư 0,02V

\(V_{saudd}=V+V=2V\)

\(\left[H^+\right]_{Dư}=\dfrac{0.02}{2}=0.01\left(M\right)\)

=>PH=2

\(n_{OH^-}=n_{NaOH}=0,01V\left(mol\right);n_{H^+}=n_{HCl}=0,03V\left(mol\right)\\ H^++OH^-\rightarrow H_2O\\ Vì:\dfrac{0,03V}{1}>\dfrac{0,01V}{1}\Rightarrow H^+dư\\ \left[H^+_{dư}\right]=\dfrac{0,03V-0,01V}{2V}=0,01\left(M\right)\\ \Rightarrow pH=-log\left[H^+_{dư}\right]=-log\left[0,01\right]=2\)

200 ml dung dịch H₂SO₄ có pH = 1 nên [ H⁺ = 0,1]

=> n_H⁺ = 2_nH2SO4 = 2 . 0,1 . 0,2 = 0,4 mol

n_OH- = n_NaOH = \(\dfrac{0,3}{40}\) = 0,0075 mol

H⁺ + OH^- -> H₂O

0,0075 <- 0,0075

=> n_H + còn = 0,04 - 0,03 = 0,01 mol

=> p_H = - lg [H⁺] = - lg (\(\dfrac{0,1}{0,2}\) ) = 0,5 mol

`1)[H^+]=0,005(M)=>pH=-log(0,005)=2,3`

`2)[OH^-]=0,002.2=4.10^[-3](M)=>pH=-log([10^[-14]]/[4.10^[-3]])=11,6`

`3)[H^+]=0,005.2=10^[-2](M)=>pH=2`

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

     `H^[+] + OH^[-] -> H_2 O`

`8,75.10^[-3]`  `8,75.10^[-3]`                     `(mol)`

`n_[H^+]=(0,0125+0,0375.2).0,1=8,75.10^[-3](mol)`

`n_[OH^-]=(0,1.2+0,1).0,4=0,12(mol)`

  `=>[OH^-]=[0,12-8,75.10^[-3]]/[0,1+0,4]=0,2225(M)`

    `=>pH=-log([10^[-14]]/[0,225])=13,35`

\(n_{H^+}=0,2.0,04.1+2.0,005.0,2=0,01\left(mol\right)\\ H^++OH^-\rightarrow H_2O\\ n_{OH^-}=n_{H^+}=0,01\left(mol\right)\\ pH=12\Rightarrow14+log\left[OH^-_{dư}\right]=12\\ \Leftrightarrow log\left[OH_{dư}^-\right]=-2\\ \Leftrightarrow\left[OH^-_{dư}\right]=0,01\left(M\right)\\ n_{OH^-_{dư}}=0,01.0,2=0,002\left(mol\right)\\ a=C_{MddNaOH}=\dfrac{0,01+0,002}{0,2}=0,06\left(M\right)\)

`OH^[-] + H^[+] -> H_2 O`

`0,008`   `0,008` 

`n_[OH^-]=0,06.2.0,1=0,012(mol)`

`n_[H^+]=0,02.0,4=0,008(mol)`

  `=>[OH^-]=[0,012-0,008]/[0,1+0,4]=0,008(M)`

  `=>pH=-log([10^[-14]]/[0,008])=11,9`

  `=>` Môi trường kiềm

\(n_{H^+}=0,1.0,04+0,005.2.0,1=0,005\left(mol\right)\\ n_{OH^-}=0,1.0,03=0,003\left(mol\right)\\ H^++OH^-\rightarrow H_2O\\ Vì:\dfrac{0,005}{1}>\dfrac{0,003}{1}\Rightarrow H^+dư\\ n_{H^+\left(Dư\right)}=0,005-0,003=0,002\left(mol\right)\\ \left[H^+_{dư}\right]=\dfrac{0,002}{0,1+0,1}=0,01\left(M\right)\\ \Rightarrow pH=-log\left[H^+_{dư}\right]=-log\left[0,01\right]=2\\ \Rightarrow Môi.trường.axit\)