Pn sửa chỗ C₃ - 0,2 thành C₃ - 3 nha, mk ghi nhằm xl, ở dưới pn cũng sửa dùm mk luôn nha, thành thật xl bạn

Sửa

\(n_{H_2SO_4\left(2M\right)}=0,15.2=0,3\left(mol\right)\)

\(n_{H_2SO_4\left(3M\right)}=0,15.3=0,45\left(mol\right)\)

\(n_{H_2SO_4\left(B\right)}=0,3+0,45=0,75\left(mol\right)\)

\(\Rightarrow C_{M_{ddB}}=\dfrac{0,75}{0,2}=3,75M\)

\(n_{H_2SO_4\left(tổng\right)}=0,15.2+0,05.3=0,45\left(mol\right)\\ V_{ddH_2SO_4\left(tổng\right)}=150+50=200\left(ml\right)=0,2\left(l\right)\\ C_{MddH_2SO_4\left(sau\right)}=C_{MddB}=\dfrac{0,45}{0,2}=2,25\left(M\right)\)

Bài 1:

Ta có: \(\Sigma n_{OH^-}=n_{NaOH}+2n_{Ba\left(OH\right)_2}=0,05.0,01+0,05.0,005.2=0,001\left(mol\right)\)

\(n_{H^+}=n_{HCl}=0,05.0,015=0,00075\left(mol\right)\)

PT ion: \(OH^-+H^+\rightarrow H_2O\)

______0,001__0,00075 (mol)

⇒ OH^- dư. n_{OH^- (dư)} = 2,5.10^-4 (mol)

\(\Rightarrow\left[OH^-\right]=\frac{2,5.10^{-4}}{0,1}=2,5.10^{-3}M\Rightarrow\left[H^+\right]=4.10^{-12}M\)

\(\Rightarrow pH\approx11,4\)

Bài 2: Đáp án D

Giải:

Ta có: \(\Sigma n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,1.0,002+0,2.2.x=2.10^{-4}+0,4x\left(mol\right)\)

\(\Rightarrow\left[H^+\right]=\frac{2.10^{-4}+0,4x}{0,3}M\)

\(\Rightarrow pH=-log\left(\frac{2.10^{-4}+0,4x}{0,3}\right)=2,7\)

\(\Rightarrow x\approx9,964.10^{-4}\approx10^{-3}\)

Bạn tham khảo nhé!

PT ion: \(H^++OH^-\rightarrow H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{H^+}=0,2\cdot0,5\cdot2=0,2\left(mol\right)\\n_{OH^-}=0,05\cdot2=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) H⁺ còn dư 0,1 mol

\(\Rightarrow\left[H^+\right]=\dfrac{0,1}{0,25}=0,4\left(M\right)\) \(\Rightarrow pH=-log\left(0,4\right)\approx0,4\)

+) Ta có:

 `V(dd(H_2SO_4)(1))=300(ml)=0,3(l)`

`V(dd(H_2SO_4)(2))=200(ml)=0,2(l)`

`=> n_((H_2SO_4)(1))=0,3.2=0,6(mol)`

 `->n_((H_2SO_4)(2))=0,2.1=0,2(mol)`

=> Nồng độ mol của dung dịch H2SO4 sau trộn:

`CM_(ddH2SO4(mới))=((0,6+0,2)/(0,3+0,2))=1,6M`

`#BTrancuato<3`

\(n_{HCl}=0,2.0,5=0,1mol\\ n_{K_2CO_3}=0,05.0,8=0,04mol\\ 2H^++CO_3^{2-}->H_2O+CO_2\\ n_{H^+dư}=0,1-0,08=0,02mol\\ C_{M\left(K^{^+}\right)}=\dfrac{0,08}{0,25}=0,32M\\ C_{M\left(H^{^+}dư\right)}=\dfrac{0,02}{0,25}=0,08M\\ C_{M\left(Cl^{^{ }-}\right)}=\dfrac{0,1}{0,25}=0,4M\)

n HNO3=0,01.0,05=5.10^-4mol

n NaNO3+ =0,015.0,1=1,5.10^-3 mol

=>CM H+= 0,01M

=>CM Na+ =0,015M

=>CmNO3- =0,01+0,015=0,025M

\(n_{H_2SO_4}=0,05.0,4=0,02\left(mol\right)\\ n_{HCl}=0,35.0,2=0,07\left(mol\right)\\ \left[H_2SO_4\right]=\dfrac{0,02}{0,05+0,35}=0,05\left(M\right)\\ \left[HCl\right]=\dfrac{0,07}{0,05+0,35}=0,175\left(M\right)\\ \Rightarrow\left[H^+\right]=0,05.2+0,175.1=0,275\left(M\right)\\ \left[SO^{2-}_4\right]=0,05\left(M\right)\\ \left[Cl^-\right]=0,175\left(M\right)\)

Đổi 200ml = 0.2 (l), 400 ml = 0.4 (l)

\(n_{HCl}=0,2.1=0,2\left(mol\right)\)

\(n_{H2SO4}=0,2.1=0,2\left(mol\right)\)

\(\Rightarrow CM_{dd\left(H2SO4\right)}=\frac{0,2}{0,4}=0,5M\)

\(\Rightarrow CM_{dd\left(H2SO4\right)}=\frac{0,2}{0,4}=0,5M\)