$n_{Ba^{2+}} = 0,1.0,5 = 0,05 < n_{SO_4^{2-}} = 0,1$ nên $SO_4^{2-}$ dư

$n_{BaSO_4} = n_{Ba^{2+}} = 0,05(mol)$
$m_{BaSO_4} = 0,05.233 = 11,65(gam)$

$n_{OH^-} = 0,1.0,5.2 + 0,1.0,5 = 0,15(mol)$
$n_{H^+} = 0,1.2 = 0,2(mol)$

$H^+ + OH^- \to H_2O$

$n_{H^+\ dư} = 0,2 - 0,15 = 0,05(mol)$

$V_{dd} = 0,1 + 0,1 + 0,1 = 0,3(lít)$

$[H^+] = \dfrac{0,05}{0,3} = \dfrac{1}{6}M$
$pH = -log( \dfrac{1}{6} ) = 0,778$

\(n_{Ba^{2+}}=0.1\cdot0.5=0.05\left(mol\right)\)

\(n_{OH^-}=0.1\cdot0.5\cdot2+0.1\cdot0.5=0.15\left(mol\right)\)

\(n_{H^+}=2\cdot0.1\cdot1=0.2\left(mol\right)\)

\(n_{SO_4^{2-}}=0.1\left(mol\right)\)

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

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

\(SO_4^{2-}dư\)

\(m_{\downarrow}=0.05\cdot233=11.65\left(g\right)\)

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

\(0.15.......0.15\)

\(n_{H^+\left(dư\right)}=0.2-0.15=0.05\left(mol\right)\)

\(\left[H^+\right]=\dfrac{0.05}{0.1+0.1+0.1}=\dfrac{1}{6}\)

\(pH=-log\left(\dfrac{1}{6}\right)=0.77\)

Đáp án B

n_H+ = 0,05 mol; n_OH- = 0,07 mol

H⁺   + OH^- → H₂O

Ta thấy H⁺ hết, OH^- dư nên nhúng quỳ tím vào Y thì quỳ tím chuyển thành màu xanh.

https://tuhoc365.vn/qa/tron-100-ml-dung-dich-a-gom-khco3-1m-va-k2co3-1m-vao-100-ml-dung-dich-b-gom-nahco3-1m-va-na2co3-1m-t/

Bạn ghi rõ từng trường hợp ra nhé!

Đáp án C

n_H+ = 0,02 mol; n_OH-= 0,04 mol

H⁺    + OH^- → H₂O

0,02   0,04

n_{OH- dư} = 0,02 mol; [OH^-] _dư = 0,02/0,2 = 0,1M suy ra [H⁺] = 10^-13 M suy ra pH = 13

Đáp án C

[OH^-]= (2.0,5.0,1+ 0,1.0,5)/ 0,2= 0,75M → [H⁺]=10^-14/ 0,75M

Nên pH= -log[H⁺]= 13,875

Ta có: X gồm \(\left\{{}\begin{matrix}HCO_3^-:0,1a\left(mol\right)\\CO_3^{2-}:0,1\left(mol\right)\end{matrix}\right.\) và Y gồm: \(\left\{{}\begin{matrix}H^+:0,2\left(mol\right)\\SO_4^{2-}:0,025\left(mol\right)\\Cl^-:0,15\left(mol\right)\end{matrix}\right.\)

- Nhỏ X vào Y:

\(HCO_3^-+H^+\rightarrow CO_2+H_2O\)

k.0,1a___k.0,1a___k.0,1a (mol)

\(CO_3^{2-}+2H^+\rightarrow CO_2+H_2O\)

k.0,1____2.k.0,1____k.0,1 (mol)

n_H⁺ = k.0,1a + 2k.0,1 = 0,2

n_CO2 = k.0,1a + k.0,1 = 0,12

\(\Rightarrow\dfrac{k.0,1a+2k.0,1}{k.0,1a+0,1k}=\dfrac{0,2}{0,12}\)

⇒ a = 0,5 (M)

- Nhỏ Y và X:

\(CO_3^{2-}+H^+\rightarrow HCO_3^-\)

0,1______0,1______0,1 (mol)

\(HCO_3^-+H^+\rightarrow CO_2+H_2O\)

__0,15___0,1_____0,1 (mol)

→ Z gồm: SO₄^2-: 0,025 (mol) và HCO₃^-: 0,05 (mol)

- Cho Ba(OH)₂ dư vào:

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

_______0,025______0,025 (mol)

\(OH^-+HCO_3^-\rightarrow CO_3^{2-}+H_2O\)

_________0,05_____0,05 (mol)

\(Ba^{2+}+CO_3^{2-}\rightarrow BaCO_3\)

________0,05_______0,05 (mol)

⇒ m_↓ = m_BaSO4 + m_BaCO3 = 15,675 (g) = m

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

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

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

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

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

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

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

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

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

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

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

Đáp án B

[OH^-]= (0,5.2.0,1+0,1.0,5)/0,2= 0,75M