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\(n_{NaOH}=\dfrac{200\cdot2\%}{40}=0.1\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{50\cdot49\%}{98}=0.25\left(mol\right)\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
\(0.1...............0.05...........0.05\)
\(n_{Na_2SO_4}=0.05\left(mol\right)\)
\(n_{H_2SO_4\left(dư\right)}=0.25-0.05=0.2\left(mol\right)\)
\(V_{dd}=\dfrac{200}{1}+\dfrac{50}{1.05}=247.6\left(ml\right)=0.2476\left(l\right)\)
\(\left[Na^+\right]=\dfrac{0.05\cdot2}{0.2476}=0.4\left(M\right)\)
\(\left[H^+\right]=\dfrac{0.2\cdot2}{0.2476}=1.6\left(M\right)\)
\(\left[SO_4^{2-}\right]=\dfrac{0.05+0.2}{0.2476}=1\left(M\right)\)
$n_{NaOH} = \dfrac{200.2\%}{40} = 0,1(mol)$
$n_{H_2SO_4} = \dfrac{50.49\%}{98} = 0,25(mol)$
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} : 2 < n_{H_2SO_4} : 1$ nên $H_2SO_4$ dư
$n_{H_2SO_4\ dư} = 0,25 - 0,1.0,5 = 0,2(mol)$
$n_{H^+\ dư} = 0,2.2 = 0,4(mol)$
Sau phản ứng :
$V_{dd} = \dfrac{200}{1} + 50.1,05 = 252,5(ml) = 0,2525(lít)$
Bảo toàn Na, S ta có :
$[Na^+] = \dfrac{0,1}{0,2525} = 0,4M$
$[SO_4^{2-}] = \dfrac{0,25}{0,2525} = 0,99M$
$[H^+] = \dfrac{0,4}{0,2525} = 1,58M$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(b.n_{NaOH\left(tổng\right)}=0,4.0,5+\dfrac{100.1,33.20\%}{40}=0,865\left(mol\right)\\ \left[Na^+\right]=\left[OH^-\right]=\left[NaOH\left(sau\right)\right]=\dfrac{0,865}{0,4+0,1}=1,73\left(M\right)\\ c.n_{HCl}=0,05.0,12=0,006\left(mol\right)\\ n_{HNO_3}=0,15.0,1=0,015\left(mol\right)\\ \left[H^+\right]=\dfrac{0,006+0,015}{0,05+0,15}=0,105\left(M\right)\\ \left[NO^-_3\right]=\dfrac{0,015}{0,05+0,15}=0,075\left(M\right)\\ \left[Cl^-\right]=\dfrac{0,006}{0,05+0,15}=0,03\left(M\right)\)
\(d.n_{H_2SO_4}=0,4.0,05=0,02\left(mol\right)\\ n_{HCl}=0,35.0,2=0,07\left(mol\right)\\ \left[H^+\right]=\dfrac{0,02.2+0,07}{0,05+0,35}=0,275\left(M\right)\\ \left[SO^{2-}_4\right]=\dfrac{0,02}{0,05+0,35}=0,05\left(M\right)\\ \left[Cl^-\right]=\dfrac{0,07}{0,05+0,35}=0,175\left(M\right)\\ f.n_{KOH}=\dfrac{20.1,31.32\%}{56}=\dfrac{131}{875}\left(mol\right)\\ n_{Ba\left(OH\right)_2}=0,08.1=0,08\left(mol\right)\\ \left[OH^-\right]=\dfrac{\dfrac{131}{875}+0,08.2}{0,02+0,08}=\dfrac{542}{175}\left(M\right)\\ \left[Ba^{2+}\right]=\dfrac{0,08}{0,02+0,08}=0,8\left(M\right)\)
\(\left[K^+\right]=\dfrac{\dfrac{131}{875}}{0,02+0,08}=\dfrac{262}{175}\left(M\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H^+}=n_{HCl}=0,03mol\)
\(n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,07mol\)
\(H^+\) + \(OH^-\) \(\rightarrow\) \(H_2O\)
bđ 0,03 0,07
pư 0,03 0,03 0,03
kt 0 0,04 0,03
\(\Rightarrow\)\(\left[OH^-\right]=\dfrac{n_{sau}}{V}=\dfrac{0,04}{1}=0,04M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(n_{H^+}=n_{OH^-}\)
\(\Leftrightarrow0,01.V_{ddHCl}=\left(0,02.2+0,01\right).0,2\)
\(\Leftrightarrow V_{ddHCl}=1\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
50 gam CuSO4.5H2O có số mol = 0.2 mol
CuSO4 > Cu(2+) + SO4(2-)
0.2----------0.2---------0.2
Cm (Cu2+) = Cm SO4(2-) = 0.2/0.2 = 1M
---------------------------------------...
pH = 2 => [H+] = 0.01M => nH+ = 0.01x0.2 = 0.002 mol
pH = 1 => [H+] = 0.1M => nH+ = 0.1x0.2 = 0.02 mol
=> nHCl cần = nH+ cần = 0.02-0.002 = 0.018 mol
---------------------------------------...
nCH3COOH = 0.01 mol
Cm dd mới pha = 0.01/0.1 = 0.1M
Ka = [CH3COO-][H+]/[CH3COOH] = 4.70 <=> a^2/(0.1-a) = 4.70 <=> a^2 = 0.47 - 4.70a <=> a = ...
Tính đc A rồi ra đc pH=-log[a]
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Dung dịch HCl 7.3% (D = 1.25g/ml)
Áp dụng công thức CM = C%*10D/M.
suy ra CM HCl = 7.3 *10*1,25/36.5= 2.5M.
Hay [HCl]= 2.5M
HCl là chất điện li mạnh trong dung dịch nên HCl --> H+ + Cl-
[H+]= [Cl-] = [HCl]= 2.5M
b) nK2SO4 = 0.022 mol.
Suy ra [K2SO4]= 0.022*0.4 = 0.0088M
K2SO4 là chất điện li mạnh nên có phương trình điện li:
K2SO4 --> 2K+ + SO4 2-
vậy [K+] = 2[K2SO4] = 0.0176M
[ SO4 2-] = [K2SO4] = 0.0088M
c) nH2SO4 = 0.24 mol.
[H2SO4] = 0.24*1.25 = 0.3M.
PT điện li : H2SO4 --> 2H+ + SO4 2-
vậy [H+] = 2[H2SO4] = 0.6M
[ SO4 2-] = [H2SO4] = 0.3M
![](https://rs.olm.vn/images/avt/0.png?1311)
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 !!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{NaOH}=0,02.2=0,04\left(mol\right)\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ a.n_{H_2SO_4}=n_{Na_2SO_4}=\dfrac{0,04}{2}=0,02\left(mol\right)\\ C_{MddH_2SO_4}=\dfrac{0,02}{0,08}=0,25\left(M\right)\\ b.\left[Na^+\right]=\dfrac{0,02.2}{0,02+0,08}=0,4\left(M\right)\\ \left[SO^{2-}_4\right]=\dfrac{0,02}{0,02+0,08}=0,2\left(M\right)\)
ta có : \(D=\dfrac{m}{V}\Leftrightarrow1,25=\dfrac{50}{V}\Rightarrow V=40\left(ml\right)\)
đổi \(40ml=0,04l\)
ta có : \(C\%=\dfrac{m_{ct}}{m_{dd}}\times100\)
\(\Rightarrow7,3=\dfrac{36,5\times n_{HCl}}{50}\Leftrightarrow n_{HCl}=0,1\left(mol\right)\)
PT ion : \(HCl\rightarrow H^++Cl^-\)
\(\Rightarrow CM_{H^+}=\dfrac{n_{H^+}}{V}=\dfrac{0,1}{0,04}=2,5M\)
\(CM_{OH^-}=\dfrac{n_{OH^-}}{V}=\dfrac{0,1}{0,04}=2,5M\)
mình dùng công thức CM = \(\dfrac{C\%.10D}{M}\)
đc k?