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\(n_{H^+}=0,2.\left(0,02+0,01.2\right)=0,008\left(mol\right)\)
\(n_{OH^-}=0,3.2.0,04=0,024\left(mol\right)\)
\(n_{OH^-dư}=0,3.2.0,04=0,016\left(mol\right)\)
\(\Rightarrow\left[OH^-_{dư}\right]=\dfrac{0,016}{0,5}=0,032M\)
\(\Rightarrow\left[H^+\right]=3,125.10^{-13}M\)
\(\Rightarrow pH\approx12,5\)
Ta có: \(\left\{{}\begin{matrix}n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,1.0,01=0,001\left(mol\right)\\n_{OH^-}=2n_{Ba\left(OH\right)_2}=2.0,1.0,01=0,002\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\Sigma n_{H^+}=0,2.0,01+2.0,2.0,02=0,01\left(mol\right)\\n_{SO_4^{2-}}=n_{H_2SO_4}=0,2.0,02=0,004\left(mol\right)\end{matrix}\right.\)
PT ion: \(H^++OH^-\rightarrow H_2O\)
_____ 0,01___0,002_________ (mol)
⇒ H+ dư. \(\Rightarrow n_{H^+\left(dư\right)}=0,008\left(mol\right)\)
\(\Rightarrow\left[H^+\right]=\frac{0,008}{0,3}=\frac{2}{75}M\Rightarrow pH\approx1,57\)
PT ion: \(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_{4\downarrow}\)
______ 0,001__0,004__ → 0,001 (mol)
\(\Rightarrow m_{\downarrow}=m_{BaSO_4}=0,001.233=0,233\left(g\right)\)
Bạn tham khảo nhé!
Ta có : \(n_{H^+}=0,3\times0,05=0,015\left(mol\right)\)
\(n_{OH^-}=2\times a\times0,2=0,4a\left(mol\right)\)
Theo bài \(PH=12\) là môi trường bazo. Vậy OH- hết
PT: \(H^++OH^-\rightarrow H_2O\)
o,4a 0,4a (mol)
\(\Rightarrow H^+dư=0,015-0,4a\)
Ta có : \(\dfrac{0,015-0,4a}{0,5}=10^{-12}\Leftrightarrow a=0,375\left(M\right)\)
Ta có: \(\left\{{}\begin{matrix}n_{H^+}=0,015\left(mol\right)\\n_{OH^-}=0,2a\left(mol\right)\end{matrix}\right.\)
Ta có: pH = 12
=> \(\dfrac{0,015-0,2a}{0,5}=10^{-12}\)
\(\Rightarrow a=0,075\)
a, \(\left\{{}\begin{matrix}n_{Ba^{2+}}=4.10^{-3}\left(mol\right)\\n_{Na^+}=3.10^{-3}\left(mol\right)\\n_{OH^-}=0,011\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[Ba^{2+}\right]=\dfrac{4.10^{-3}}{0,2+0,3}=0,008M\\\left[Na^+\right]=\dfrac{3.10^{-3}}{0,2+0,3}=0,006M\\\left[OH^-\right]=\dfrac{0,011}{0,2+0,3}=0,022M\end{matrix}\right.\)
b, Để trung hòa dung dịch A thì:
\(n_{H^+}=n_{OH^-}\)
\(\Leftrightarrow0,01.V_{ddHCl}=\left(0,02.2+0,01\right).0,2\)
\(\Leftrightarrow V_{ddHCl}=1\left(l\right)\)
Bài 1:
Ta có: \(\Sigma n_{OH^-}=n_{NaOH}+2n_{Ba\left(OH\right)_2}=0,05.0,01+0,05.0,005.2=0,001\left(mol\right)\)
\(n_{H^+}=n_{HCl}=0,05.0,015=0,00075\left(mol\right)\)
PT ion: \(OH^-+H^+\rightarrow H_2O\)
______0,001__0,00075 (mol)
⇒ OH- dư. nOH- (dư) = 2,5.10-4 (mol)
\(\Rightarrow\left[OH^-\right]=\frac{2,5.10^{-4}}{0,1}=2,5.10^{-3}M\Rightarrow\left[H^+\right]=4.10^{-12}M\)
\(\Rightarrow pH\approx11,4\)
Bài 2: Đáp án D
Giải:
Ta có: \(\Sigma n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,1.0,002+0,2.2.x=2.10^{-4}+0,4x\left(mol\right)\)
\(\Rightarrow\left[H^+\right]=\frac{2.10^{-4}+0,4x}{0,3}M\)
\(\Rightarrow pH=-log\left(\frac{2.10^{-4}+0,4x}{0,3}\right)=2,7\)
\(\Rightarrow x\approx9,964.10^{-4}\approx10^{-3}\)
Bạn tham khảo nhé!
