\(b.n_{NaOH\left(tổng\right)}=0,4.0,5+\dfrac{100.1,33.20\%}{40}=0,865\left(mol\right)\\ \left[Na^+\right]=\left[OH^-\right]=\left[NaOH\left(sau\right)\right]=\dfrac{0,865}{0,4+0,1}=1,73\left(M\right)\\ c.n_{HCl}=0,05.0,12=0,006\left(mol\right)\\ n_{HNO_3}=0,15.0,1=0,015\left(mol\right)\\ \left[H^+\right]=\dfrac{0,006+0,015}{0,05+0,15}=0,105\left(M\right)\\ \left[NO^-_3\right]=\dfrac{0,015}{0,05+0,15}=0,075\left(M\right)\\ \left[Cl^-\right]=\dfrac{0,006}{0,05+0,15}=0,03\left(M\right)\)

\(d.n_{H_2SO_4}=0,4.0,05=0,02\left(mol\right)\\ n_{HCl}=0,35.0,2=0,07\left(mol\right)\\ \left[H^+\right]=\dfrac{0,02.2+0,07}{0,05+0,35}=0,275\left(M\right)\\ \left[SO^{2-}_4\right]=\dfrac{0,02}{0,05+0,35}=0,05\left(M\right)\\ \left[Cl^-\right]=\dfrac{0,07}{0,05+0,35}=0,175\left(M\right)\\ f.n_{KOH}=\dfrac{20.1,31.32\%}{56}=\dfrac{131}{875}\left(mol\right)\\ n_{Ba\left(OH\right)_2}=0,08.1=0,08\left(mol\right)\\ \left[OH^-\right]=\dfrac{\dfrac{131}{875}+0,08.2}{0,02+0,08}=\dfrac{542}{175}\left(M\right)\\ \left[Ba^{2+}\right]=\dfrac{0,08}{0,02+0,08}=0,8\left(M\right)\)

\(\left[K^+\right]=\dfrac{\dfrac{131}{875}}{0,02+0,08}=\dfrac{262}{175}\left(M\right)\)

Đề có cho thể tích dd H₂SO₄ không bạn nhỉ?

PTHH: \(H_2SO_4+2KOH\rightarrow K_2SO+2H_2O\)

Ta có: \(n_{H_2SO_4}=0,2\cdot1=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{KOH}=0,4\left(mol\right)\\n_{K_2SO_4}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{ddKOH}=\dfrac{\dfrac{0,4\cdot56}{6\%}}{1,048}\approx356,2\left(ml\right)\\C_{M_{K_2SO_4}}=\dfrac{0,2}{0,2+0,3562}\approx0,36\left(M\right)\end{matrix}\right.\)   

\(n_{H_2SO_4}=1.0,2=0,2\left(mol\right)\\ H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\\ 0,2.........0,4........0,2.......0,2\left(mol\right)\\ a.m_{ddKOH}=\dfrac{0,4.56.100}{6}=\dfrac{1120}{3}\left(g\right)\\ V_{ddKOH}=\dfrac{\dfrac{1120}{3}}{1,048}=\dfrac{140000}{393}\left(ml\right)\approx0,356\left(l\right)\)

\(b.C_{MddK_2SO_4}=\dfrac{0,2}{\dfrac{140000}{393}+0,2}\approx0,00056\left(M\right)\)

2

b

mNaCl=\(\dfrac{200.15}{100}\)=30(g)

nNaCl=\(\dfrac{30}{58,5}\)=0.51(mol)

VddNaCl=\(\dfrac{200}{1,1}\)=181.8(ml)=0.1818(l)

CMNaCl=\(\dfrac{0,51}{0,1818}\)=2.8(M)

\(n_{HCl.5M}=0,05\times5=0,25\left(mol\right)\)

\(m_{ddHCl.30\%}=200\times1,33=266\left(g\right)\)

\(\Rightarrow m_{HCl.30\%}=266\times30\%=79,8\left(g\right)\)

\(\Rightarrow n_{HCl.30\%}=\frac{79,8}{36,5}=\frac{798}{365}\left(mol\right)\)

\(\Rightarrow\Sigma n_{HCl}mới=0,25+\frac{798}{365}=\frac{3557}{1460}\left(mol\right)\)

\(\Sigma V_{ddHCl}mới=50+200=250\left(ml\right)=0,25\left(l\right)\)

\(\Rightarrow C_{M_{HCl}}mới=\frac{3557}{1460}\div0,25=9,75\left(M\right)\)

1. Bài này câu b hình như tính C% chứ nhỉ ?

---------------

\(n_{Na_2CO_3}=\dfrac{100.10,6\%}{106}=0,1\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{200.9,8\%}{98}=0,2\left(mol\right)\)

Pt: \(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2\uparrow+H_2O\)

0,1mol 0,2mol ---------> 0,1mol----> 0,1mol

Lập tỉ số: \(n_{Na_2CO_3}:n_{H_2SO_4}=0,1< 0,2\)

=> \(Na_2CO_3\) hết, H₂SO₄ dư

\(V_{CO_2}=0,1.22,4=2,24\left(l\right)\)

\(\Sigma_{m_{dd}\left(spu\right)}=100+200-0,1.44=295,6\left(g\right)\)

\(C\%_{Na_2SO_4}=\dfrac{0,1.82.100}{295,6}=2,77\%\)

\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(0,2-0,1\right).98.100}{295,6}=3,32\%\)

2. Ở câu b: Nếu đổi thành dd KOH 5% KLR là 1,045g/ml thì cần bao nhiêu ml dd KOH.

KOH chứ không phải NaOH. Lần sau ghi đề chú ý dùm.

\(n_{H_2SO_4}=1,5.0,05=0,075\left(mol\right)\)

\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

0,075mol --> 0,15mol

\(m_{dd}=\dfrac{0,15.40.100}{30}=20\left(g\right)\)

b) \(H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\)

0,075mol----> 0,15mol

\(m_{dd}=\dfrac{0,15.56.100}{5}=168\left(g\right)\)

\(V_{KOH}=\dfrac{168}{1,045}=160,76\left(ml\right)\)

3. \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,25mol -----------> 0,5mol

\(C_{M_{NaOH}}=\dfrac{0,5}{0,5}=1M\)

\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

0,25mol<--- 0,5mol

\(m_{dd_{H_2SO_4}}=\dfrac{0,25.98.100}{20}=122,5\left(g\right)\)

\(V_{H_2SO_4}=\dfrac{122,5}{1,14}=107,46\left(ml\right)\)