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

\(\left[OH^-\right]=0,001M\\ \Rightarrow pH=14-pOH=14+log\left[OH^-\right]=14+log\left[0,001\right]=14-3=11\)

Ta có: \(n_{H^+}=n_{HCl}=0,04.0,75=0,03\left(mol\right)\)

\(n_{OH^-}=2n_{Ba\left(OH\right)_2}+n_{KOH}=2.0,16.0,08+0,16.0,04=0,032\left(mol\right)\)

PT: \(H^++OH^-\rightarrow H_2O\)

____0,03______0,03 (mol)

\(\Rightarrow n_{OH^-\left(dư\right)}=0,032-0,03=0,002\left(mol\right)\)

\(\Rightarrow\left[OH^-\right]=\dfrac{0,002}{0,04+0,16}=0,01\left(M\right)\)

\(\Rightarrow\left[H^+\right]=\dfrac{10^{-14}}{0,01}=10^{-12}\left(M\right)\)

\(\Rightarrow pH=-log\left[H^+\right]=12\)

Ta có: \(\left\{{}\begin{matrix}n_{H_2SO_4}=\dfrac{0,098}{98}=0,001\left(mol\right)\\n_{HNO_3}=0,2.0,04=0,008\left(mol\right)\end{matrix}\right.\)

=> \(\sum n_{H^+}=2n_{H_2SO_4}+n_{HNO_3}=0,01\left(mol\right)\)

=> \(\left[H^+\right]=\dfrac{0,01}{0,5+0,2}=\dfrac{1}{70}M\)

=> \(pH=-\log\left(\dfrac{1}{70}\right)=1,845\)

Ta có: \(n_{KOH}=n_{H_2SO_4}=0,25\left(mol\right)\)

PT: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

Xét tỉ lệ: \(\dfrac{0,25}{2}< \dfrac{0,25}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=\dfrac{1}{2}n_{KOH}=0,125\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,25-0,125=0,125\left(mol\right)\)

\(\Rightarrow n_{H^+}=0,125.2=0,25\left(mol\right)\)

\(\Rightarrow\left[H^+\right]=\dfrac{0,25}{0,25+0,25}=0,5\left(M\right)\)

\(\Rightarrow pH=-\log\left(0,5\right)\approx0,3\)

→ Vậy: Dd sau pư có MT axit.

\(n_{OH^-}=0,15.10^{-2}\)

\(n_{H^+}=0,35.\left(0,02+0,01.2\right)=0,014\)

\(n_{H^+d\text{ư}}=0,014-0,15.10^{-2}=0,0125\Rightarrow\left[H^+\right]d\text{ư}=\dfrac{0,0125}{0,15+0,35}=0,025\)

\(\Rightarrow pH\approx1,6\)

Ta có: \(\left\{{}\begin{matrix}n_{H^+}=n_{HNO_3}=\dfrac{20.31,5\%}{63}=0,1\left(mol\right)\\n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,5.0,2.2=0,2\left(mol\right)\end{matrix}\right.\)

PTHH: \(H^++OH^-\rightarrow H_2O\)

           0,1--->0,1

`=>` \(n_{OH^-\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

Đề thiếu V_{dd HNO3} nên không tính được V_{dd sau phản ứng} , không tìm được pH

\(n_{H_2SO_4}=\dfrac{2,45}{98}=0,025\left(mol\right)\Rightarrow n_{H^+}=0,025\cdot2=0,05\left(mol\right)\)

\(n_{NaOH}=0,3\cdot0,15=0,045=n_{OH^-}\)

PT ion RG: \(H^++OH^-\rightarrow H_2O\)

                 0,045 ← 0,045

\(\Rightarrow n_{H^+\text{ (dư)}}=0,05-0,045=0,005\left(mol\right)\)

\(\Rightarrow\left[H^+\right]_{\text{(dư)}}=\dfrac{0,005}{0,35+0,15}=0,01\left(M\right) \)

\(\Rightarrow pH=-\log0,01=2\)

a) \(\left[OH^-\right]=0,04+0,06=0,1M=10^{-1}M\)

`=>` \(pOH=1\Rightarrow pH=14-1=13\)

b) \(\left[H^+\right]=0,05.2+0,1=0,2M\)

`=>` \(pH=-log\left(0,2\right)=0,7\)

a) \(\Sigma\left[OH^-\right]=0,1M\)

\(\Rightarrow\left[H^+\right]=\dfrac{10^{-14}}{0,1}=10^{-13}\Rightarrow pH=13\)

b) \(\Sigma\left[H^+\right]=0,2M\)

\(\Rightarrow pH=-\log\left(0,2\right)=0,7\)

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