Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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!
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:
\(n_{H^+}=n_{OH^-}\)
\(\Leftrightarrow0,01.V_{ddHCl}=\left(0,02.2+0,01\right).0,2\)
\(\Leftrightarrow V_{ddHCl}=1\left(l\right)\)
\(n_{H^+}=0,2.\left(0,02+0,01.2\right)=0,008\left(mol\right)\)
\(n_{OH^-}=0,3.2.0,04=0,024\left(mol\right)\)
\(n_{OH^-dư}=0,3.2.0,04=0,016\left(mol\right)\)
\(\Rightarrow\left[OH^-_{dư}\right]=\dfrac{0,016}{0,5}=0,032M\)
\(\Rightarrow\left[H^+\right]=3,125.10^{-13}M\)
\(\Rightarrow pH\approx12,5\)