@Trần Hữu Tuyển help em bài này nữaaaa

@Gia Hân Ngô help tauuuu

\(n_{KOH}=0.1\cdot1=0.1\left(mol\right)\)

\(n_{H_2SO_4}=0.3\cdot0.5=0.15\left(mol\right)\)

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

\(0.1..........0.05...............0.05\)

Dung dịch D : 0.05 (mol) K₂SO₄ , 0.1 (mol) H₂SO₄

\(\left[K^+\right]=\dfrac{0.05\cdot2}{0.1+0.3}=0.25\left(M\right)\)

\(\left[H^+\right]=\dfrac{0.1\cdot2}{0.1+0.3}=0.5\left(M\right)\)

\(\left[SO_4^{2-}\right]=\dfrac{0.05+0.1}{0.1+0.3}=0.375\left(M\right)\)

\(2NaOH+H_2SO_4\rightarrow K_2SO_4+H_2O\)

\(0.2..................0.1\)

\(V_{dd_{NaOH}}=\dfrac{0.2}{1}=0.2\left(l\right)\)

\(n_{KOH}=0.1\cdot1=0.1\left(mol\right)\)

\(n_{HCl}=0.1\cdot0.5=0.05\left(mol\right)\)

\(KOH+HCl\rightarrow KCl+H_2O\)

\(0.05.......0.05.......0.05\)

Dung dịch D : 0.05 (mol) KOH , 0.05 (mol) KCl

\(\left[K^+\right]=\dfrac{0.05+0.05}{0.1+0.1}=0.5\left(M\right)\)

\(\left[Cl^-\right]=\dfrac{0.05}{0.2}=0.25\left(M\right)\)

\(\left[OH^-\right]=\dfrac{0.05}{0.2}=0.25\left(M\right)\)

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

\(0.05.........0.025\)

\(V_{dd_{H_2SO_4}}=\dfrac{0.025}{1}=0.025\left(l\right)\)

a) \(n_{NaOH}=0,2.1=0,2\left(mol\right)\); \(n_{HNO_3}=0,2.0,5=0,1\left(mol\right)\)

\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)

0,2.............0,1

Lập tỉ lệ : \(\dfrac{0,2}{1}>\dfrac{0,1}{1}\) => Sau phản ứng NaOH dư

Dung dịch D gồm NaNO3 và NaOH dư

\(n_{NaNO_3}=n_{HNO_3}=0,1\left(mol\right)\)

\(n_{NaOH\left(pứ\right)}=n_{HNO_3}=0,1\left(mol\right)\)

\(n_{NaOH\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

Ion trong dung dịch D : Na+ , NO3-, OH-

\(\left[Na^+\right]=\dfrac{0,1+0,1}{0,2}=1M\)

\(\left[NO_3^-\right]=\dfrac{0,1}{0,2}=0,5M\)

\(\left[OH^-\right]=\dfrac{0,1}{0,2}=0,5M\)

b)Trong dung dịch D chỉ có NaOH dư phản ứng

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

0,1................0,05

=> \(V_{H_2SO_4}=\dfrac{0,05}{1}=0,05\left(l\right)\)

Tách ra lần bài đi em !

100ml = 0,1l

\(n_{H2SO4}=3.0,1=0,3\left(mol\right)\)

a) Pt : \(H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O|\)

                1             2                1              2

              0,3           0,6             0,3

b) \(n_{K2SO4}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

⇒ \(m_{K2SO4}=0,3.174=52,2\left(g\right)\)

c) \(n_{KOH}=\dfrac{0,3.2}{1}=0,6\left(mol\right)\)

\(V_{ddKOH}=\dfrac{0,6}{2}=0,3\left(l\right)\)

d) \(V_{ddspu}=0,1+0,3=0,4\left(l\right)\)

\(C_{M_{K2SO4}}=\dfrac{0,3}{0,4}=0,75\left(M\right)\)

 Chúc bạn học tốt

a, Ta có: \(n_{H_2SO_4}=0,3.0,5=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{H^+}=2n_{H_2SO_4}=0,3\left(mol\right)\\n_{SO_4^{2-}}=n_{H_2SO_4}=0,15\left(mol\right)\end{matrix}\right.\)

\(n_{K^+}=n_{OH^-}=n_{KOH}=0,2.1=0,2\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

0,3 ___ 0,2 __________ (mol)

\(\Rightarrow n_{H^+\left(dư\right)}=0,1\left(mol\right)\)

⇒ Dung dịch A gồm: H⁺; SO₄^2- và K⁺

\(\Rightarrow\left\{{}\begin{matrix}\left[H^+\right]=\frac{0,1}{0,5}=0,2M\\\left[SO_4^{2-}\right]=\frac{0,15}{0,5}=0,3M\\\left[K^+\right]=\frac{0,2}{0,5}=0,4M\end{matrix}\right.\)

b, \(H^++OH^-\rightarrow H_2O\)

__0,1 → 0,1 ___________ (mol)

\(\Rightarrow n_{NaOH}=n_{OH^-}=0,1\left(mol\right)\)

\(\Rightarrow V_{NaOH}=\frac{0,1}{0,5}=0,2\left(l\right)\)

Bạn tham khảo nhé!