Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(\left[Ca^{2+}\right]=\dfrac{0,15.0,5}{0,15+0,05}=0,375M\)
\(\left[Na^+\right]=\dfrac{0,05.2}{0,15+0,05}=0,5M\)
\(\left[Cl^-\right]=\dfrac{0,15.2.0,5+0,05.2}{0,15+0,05}=1,25M\)
a) Ta có: \(n_{Al\left(NO_3\right)_3}=\dfrac{4,26}{213}=0,02\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al^+}=0,02\left(mol\right)\\n_{NO_3^-}=0,06\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[Al^+\right]=\dfrac{0,02}{0,1}=0,2\left(M\right)\\\left[NO_3^-\right]=\dfrac{0,06}{0,1}=0,6\left(M\right)\end{matrix}\right.\)
b) Ta có: \(\left[Na^+\right]=0,1+0,02\cdot2+0,3=0,304\left(M\right)\)
c) Bạn xem lại đề !!
\(n_{OH^-}=0,5.0,2+0,2.2.0,3=0,22\left(mol\right)\Rightarrow\left[OH^-\right]=\dfrac{0,22}{0,5}=0,44M\)
\(n_{Na^+}=0,5.0,2=0,1\left(mol\right)\Rightarrow\left[Na^+\right]=\dfrac{0,1}{0,5}=0,2M\)
\(n_{Ba^{2+}}=0,2.0,3=0,06\left(mol\right)\Rightarrow\left[Ba^{2+}\right]=\dfrac{0,06}{0,5}=0,12M\)
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:
\(n_{H^+}=n_{OH^-}\)
\(\Leftrightarrow0,01.V_{ddHCl}=\left(0,02.2+0,01\right).0,2\)
\(\Leftrightarrow V_{ddHCl}=1\left(l\right)\)
\(NaOH\rightarrow Na^++OH^-\)
\(Ba\left(OH\right)_2\rightarrow Ba^{2+}+2OH^-\)
\(n_{Na^+}=0,1.0,1=0,01\left(mol\right)\)
\(n_{Ba^{2+}}=0,05.0,1=0,005\left(mol\right)\)
\(n_{OH^-}=0,1.0,1+0,05.2.0,1=0,02\left(mol\right)\)
Câu 1:
a) \(Pb\left(NO_3\right)_2+CaCl_2\rightarrow PbCl_2\downarrow+Ca\left(NO_3\right)_2\)
\(Pb^{2+}+2Cl^-\rightarrow PbCl_2\downarrow\)
b) \(\left(NH_4\right)_2CO_3+2KOH\rightarrow K_2CO_3+2NH_3\uparrow+H_2O\)
\(NH_4^++OH^-\rightarrow NH_3\uparrow+H_2O\)
Câu 2:
a) Ta có:
\(\left\{{}\begin{matrix}n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,08.0,25.2=0,04\left(mol\right)\\n_{H^+}=n_{HNO_3}+2n_{H_2SO_4}=0,5.0,02+0,02.0,5.2=0,03\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,08.0,25=0,02\left(mol\right)\\n_{SO_4^{2-}}=0,02.0,5=0,01\left(mol\right)\end{matrix}\right.\)
PTHH: \(H^++OH^-\rightarrow H_2O\)
0,03--->0,03
\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\downarrow\)
0,01<---0,01
`=>` \(ddA:\left\{{}\begin{matrix}n_{OH^-\left(dư\right)}=0,04-0,03=0,01\left(mol\right)\\n_{Ba^{2+}\left(dư\right)}=0,02-0,01=0,01\left(mol\right)\\n_{NO_3^-}=n_{HNO_3}=0,02.0,5=0,01\left(mol\right)\end{matrix}\right.\)
Ta có: \(V_{ddA}=0,08+0,02=0,1\left(l\right)\)
`=>` \(\left[OH^-\right]=\left[Ba^{2+}\right]=\left[NO_3^-\right]=\dfrac{0,01}{0,1}=0,1M\)
b) \(pOH=-\text{log}\left(0,1\right)=1\)
`=>` \(pH=14-1=13\)