[K+]=[Cl-]=0,25M

[KOH dư]=0,25M

b)   2KCl   +    H₂SO₄    ----------->K₂SO₄   + 2HCl

  0,05(mol)---->0,025(mol)

=>v_H2SO4=\(\frac{0,025}{1}\)=0,025(lít)

c)pH=-log(0,25)=0,602

(câu c mình không chắc chắn lắm nha bạn!!!)

Cho mình hỏi s [K+]=[Cl-]=0,25M đc z. Mình chưa hiểu lắm

a, \(n_{HCl}=0,2.0,1=0,02\left(mol\right)=n_{H^+}=n_{Cl^-}\)

\(n_{H_2SO_4}=0,2.0,15=0,03\left(mol\right)=n_{SO_4^{2-}}\) \(\Rightarrow n_{H^+}=2n_{H_2SO_4}=0,06\left(mol\right)\)

\(\Rightarrow\Sigma n_{H^+}=0,02+0,06=0,08\left(mol\right)\)

\(n_{Ba\left(OH\right)_2}=0,3.0,05=0,015\left(mol\right)=n_{Ba^{2+}}\) 

\(\Rightarrow n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,03\left(mol\right)\)

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

0,03___0,03 (mol) ⇒ n_H⁺ dư = 0,05 (mol)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)

0,015___0,015______0,015 (mol) ⇒ n_SO4^2- dư = 0,015 (mol)

⇒ m = m_BaSO4 = 0,015.233 = 3,495 (g)

\(\left[Cl^-\right]=\dfrac{0,02}{0,2+0,3}=0,04\left(M\right)\)

\(\left[H^+\right]=\dfrac{0,05}{0,2+0,3}=0,1\left(M\right)\)

\(\left[SO_4^{2-}\right]=\dfrac{0,015}{0,2+0,3}=0,03\left(M\right)\)

b, pH = -log[H⁺] = 1

a)

Coi V dd HCl = 100(ml)

m dd HCl = 1,25.100 = 125(gam)

n HCl = 125.7,3%/36,5 = 0,25(mol)

[H⁺ ] = [Cl^- ] = CM HCl = 0,25/0,1 = 2,5M

b)

n Al = 0,235(mol)

2Al + 6HCl $\to$ 2AlCl3 + 3H2

n HCl pư = 3n Al = 0,705(mol)

n HCl dư = 0,4.2 - 0,705 = 0,095(mol)

[H⁺ ] = CM HCl dư = 0,095/0,4 = 0,2375M

pH = -log([H⁺ ]) = 0,624

a, \(\left[Na^+\right]=0,1\)

\(\left[K^+\right]=0,1\)

\(\left[OH^-\right]=0,2\)

\(\left[SO_4^{2-}\right]=0,2\)

\(\left[H^+\right]=0,4\)

b, \(n_{H^+}=0,1.0,4=0,04\left(mol\right)\)

\(n_{OH^-}=0,1.0,2=0,02\left(mol\right)\)

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

\(\Rightarrow n_{H^+dư}=0,02\left(mol\right)\)

\(\Rightarrow\left[H^+\right]=\dfrac{0,02}{200}=10^{-4}\)

\(\Rightarrow pH=4\)

$n_{NaOH} = n_{KOH} = 0,1.0,1 = 0,01(mol)$
$n_{H_2SO_4} = 0,02(mol)$

              OH^- + H⁺ → H₂O

       Bđ : 0,01...0,04..................(mol)

      Pư : 0,01...0,01...................(mol)

Sau pư :   0......0,03...................(mol)

$V_{dd} = 0,1 + 0,1 = 0,2(lít)$

Vậy : 

 $[K^+] = [Na^+] = \dfrac{0,01}{0,2} = 0,05M$
$[H^+] = \dfrac{0,03}{0,2} = 0,15M$
$[SO_4^{2-}] = \dfrac{0,02}{0,2} = 0,1M$

b)

$pH = -log(0,15) = 0,824$

a, \(n_{Ba\left(OH\right)_2}=0,1.0,1=0,01\left(mol\right)=n_{Ba^{2+}}\)

\(\Rightarrow n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,02\left(mol\right)\)

\(n_{NaOH}=0,1.0,1=0,01\left(mol\right)=n_{Na^+}=n_{OH^-}\)

⇒ Σn_OH^- = 0,02 + 0,01 = 0,03 (mol)

\(n_{H_2SO_4}=0,4.0,0175=0,007\left(mol\right)=n_{SO_4^{2-}}\)

\(\Rightarrow n_{H^+}=2n_{H_2SO_4}=0,014\left(mol\right)\)

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

0,014___0,014 (mol) ⇒ n_OH^- dư = 0,03 - 0,014 = 0,016 (mol)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)

0,007____0,007_____0,007 (mol) ⇒ n_Ba²⁺ dư = 0,01 - 0,007 = 0,003 (mol)

⇒ m = 0,007.233 = 1,631 (g)

\(\left[OH^-\right]=\dfrac{0,016}{0,1+0,4}=0,032\left(M\right)\)

\(\left[Ba^{2+}\right]=\dfrac{0,003}{0,1+0,4}=0,006\left(M\right)\)

\(\left[Na^+\right]=\dfrac{0,01}{0,1+0,4}=0,02\left(M\right)\)

b, pH = 14 - (-log[OH^-]) ≃ 12,505

\(n_{Ba^{2+}}=0,1.0,1=0,01\left(mol\right)\)

\(n_{SO_4^{2-}}=0,4.0,0175=7.10 ^{-3}\left(mol\right)\)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\downarrow\)

\(\Rightarrow m=m_{BaSO_4}=7.10^{-3}.233=1,631\left(g\right)\)

Ta có:

\(n_{H^+}=0,4.0,0175.2=0,014\left(mol\right)\)

\(n_{OH^-}=0,1.0,1.2+0,1.0,1=0,03\left(mol\right)\)

Trong dung dịch X:

\(n_{OH^-}=0,03-0,014=0,016\left(mol\right)\)\(\Rightarrow\left[OH^-\right]=\dfrac{0,016}{0,1+0,4}=0,032\left(M\right)\)

\(n_{Ba^{2+}}=0,01-7.10^{-3}=3.10^{-3}\left(mol\right)\Rightarrow\left[Ba^{2+}\right]=\dfrac{3.10^{-3}}{0,1+0,4}=6.10^{-3}\left(M\right)\)

\(n_{Na^+}=0,1.0,1=0,01\left(mol\right)\Rightarrow\left[Na^+\right]=0,02\)

\(pOH=-lg\left(0,032\right)\approx1,5\Rightarrow pH=14-1,5=12,5\)

a, \(n_{HCl}=0,1.0,2=0,02\left(mol\right)=n_{H^+}=n_{Cl^-}\)

\(n_{H_2SO_4}=0,1.0,2=0,02\left(mol\right)=n_{SO_4^{2-}}\) \(\Rightarrow n_{H^+}=2n_{H_2SO_4}=0,04\left(mol\right)\)

\(n_{NaOH}=0,3.0,4=0,12\left(mol\right)=n_{Na^+}=n_{OH^-}\)

\(\Rightarrow\sum n_{H^+}=0,02+0,04=0,06\left(mol\right)\)

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

0,06__0,06 (mol)

⇒ n_OH^- dư = 0,12 - 0,06 = 0,06 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\left[Cl^-\right]=\dfrac{0,02}{0,1+0,3}=0,05\left(M\right)\\\left[SO_4^{2-}\right]=\dfrac{0,02}{0,1+0,3}=0,05\left(M\right)\\\left[Na^+\right]=\dfrac{0,12}{0,1+0,3}=0,3\left(M\right)\\\left[OH^-\right]=\dfrac{0,06}{0,1+0,3}=0,15\left(M\right)\end{matrix}\right.\)

b, pH = 14 - (-log[OH^-]) ≃ 13,176

\(n_{OH^-}=n_{NaOH}=0,3.1,5=0,45\left(mol\right)\\ n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,2x+0,5.0,2.2=0,2x+0,2\left(mol\right)\)

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

                      0,45<---0,45

\(\Rightarrow0,2x+0,2=0,45\Leftrightarrow x=1,25M\)

Ta có: \(V_{dd}=0,3+0,2=0,5\left(l\right)\) và \(\left\{{}\begin{matrix}n_{Na^+}=0,45\left(mol\right)\\n_{Cl^-}=0,2.1,25=0,25\left(mol\right)\\n_{SO_4^{2-}}=0,2.0,5=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{Na^+}=\dfrac{0,45}{0,5}=0,9M\\C_{Cl^-}=\dfrac{0,25}{0,5}=0,5M\\C_{SO_4^{2-}}=\dfrac{0,1}{0,5}=0,2M\end{matrix}\right.\)