Bài tập về tính pH của dung dịch acid mạnh, base mạnh lớp 11 (có lời giải)

a. Đúng. [H⁺] = \frac{0,0001}{0,1} = 0,001 M = 10^-3 M.

b. Đúng. pH = -log[H⁺] = -log(0,001) = 3.

c. Sai.[H⁺] = 0,001 M trong dịch vị của bệnh nhân thấp hơn bình thường (khoảng 0,04M)

d. Sai. Phản ứng: OH^- + H⁺ → H₂O. {\text{n}}_{\text{O}{\text{H}}^{-}}=0,00005<{\text{n}}_{{\text{H}}^{+}}=0,0001 nên không trung hòa được hoàn toàn acid.

a. Sai. pH = 10,5 → [H+] =10^-10,5M → [OH^–] = 10^–3,5 M.

b. Sai.{\text{n}}_{\text{O}{\text{H}}^{-}}=0,1.10^{-3,5}=10^{-4,5}\text{(}\text{m}\text{o}\text{l}\text{)}.

c. Đúng. pOH = –log[OH^–] = –log(10^–3,5) = 3,5.

d. Đúng. Vì mẫu nước có pH > 7 nên có môi trường kiềm → làm quỳ tím chuyển xanh.

a. Đúng. {\text{n}}_{\text{K}\text{O}\text{H}}=\frac{0,56}{56}=0,01\text{(}\text{m}\text{o}\text{l}\text{)}

b. Đúng. [OH^–] = \frac{0,01}{1}= 0,01 M.

c. Sai. pH = 14 + log[OH^–] = 14 +log(0,01) = 12.

d. Đúng. Dung dịch E là dung dịch kiềm mạnh nên làm phenolphtalein chuyển màu hồng.

a. Đúng. {\text{n}}_{{\text{H}}^{+}}={\text{n}}_{\text{H}\text{C}\text{l}}=0,3.0,05=0,015\text{(}\text{m}\text{o}\text{l}\text{)}

b. Đúng. {\text{n}}_{{\text{H}}^{+}}=2{\text{n}}_{{\text{H}}_2\text{S}{\text{O}}_4}=2.0,2.0,075=0,03\text{(}\text{m}\text{o}\text{l}\text{)}

c. Đúng.{\displaystyle \sum _{}^{}{\text{n}}_{{\text{H}}^{+}}=0,015+0,03}=0,045\text{m}\text{o}\text{l}\text{;}{\text{V}}_{\text{d}\text{d}\text{s}\text{a}\text{u}}=0,2+0,3=0,5(\text{L}\text{)}\text{.}

[{\text{H}}^{+}]=\frac{0,045}{0,5}=0,09\text{M}\text{.}

d. Đúng. pH = –log(0,09) = 1,046 ≈ 1.

a. Đúng.{\displaystyle \sum _{}^{}{\text{n}}_{\text{O}{\text{H}}^{-}}=\text{2}{\text{n}}_{\text{B}\text{a}{\text{(}\text{O}\text{H}\text{)}}_{\text{2}}}\text{+}}{\text{n}}_{\text{N}\text{a}\text{O}\text{H}}=2.0,2.0,025+0,3.0,05=0,025(\text{m}\text{o}\text{l})

b. Đúng. V_dd= 0,2 + 0,3 = 0,5 (Lít). [OH^–] = \frac{0,025}{0,5}=0,05\text{M}

c. Đúng. pH = 14 + log[OH^–] = 14 +log(0,05) = 12,699 ≈ 12,7.

d. Sai. Dung dịch T có tính kiềm làm quỳ tím chuyển màu xanh.

Đáp án: 4

{\text{n}}_{{\text{H}}^{\text{+}}}\text{=}{\text{n}}_{\text{H}\text{N}{\text{O}}_{\text{3}}}\to [H^{+}]=C_{M\text{(}\text{H}\text{N}{\text{O}}_3)}=0,0001\text{M}

pH = –log[H⁺] = –log(0,0001) = 4.

Đáp án: 12

{\text{n}}_{\text{N}{\text{a}}_{\text{2}}\text{O}}=\frac{0,31}{62}=0,005\text{(}\text{m}\text{o}\text{l}\text{)}

PTHH: Na₂O + H₂O → 2NaOH

Theo pthh: {\text{n}}_{\text{N}\text{a}\text{O}\text{H}}=2{\text{n}}_{\text{N}{\text{a}}_{\text{2}}\text{O}}=2.0,005=0,01\text{(}\text{m}\text{o}\text{l}\text{)}→ {\text{n}}_{\text{O}{\text{H}}^{-}}={\text{n}}_{\text{N}\text{a}\text{O}\text{H}}=0,01\text{(}\text{m}\text{o}\text{l}\text{)}

[OH^–] = \frac{0,01}{1}= 0,01 M → pH = 14 + log[OH^–] = 14 + log(0,01) = 12.

Đáp án: 1,7

{\displaystyle \sum _{}^{}{\text{n}}_{{\text{H}}^{\text{+}}}}\text{=}\text{2}{\text{n}}_{{\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}}+{\text{n}}_{\text{H}\text{C}\text{l}}=2.0,1.0,02+0,4.0,015=0,01\text{m}\text{o}\text{l}.

$V{}_{\text{d}\text{d}}=0,1+0,4=0,5\text{L}\Rightarrow \left[\text{H}{}^{+}\right]=\frac{0,01}{0,5}=0,02\text{M}$

$\text{p}\text{H}=-log\left(0,02\right)\approx 1,70$

Đáp án: 0,08

[OH^–] = 10^{–(14 – pH)} = 10^–2,7 → {\text{n}}_{\text{O}{\text{H}}^{\text{-}}}=1.10^{-2,7}=10^{-2,7}\text{(}\text{m}\text{o}\text{l}\text{)}\text{.}

{\text{n}}_{\text{N}\text{a}\text{O}\text{H}}={\text{n}}_{\text{O}{\text{H}}^{\text{-}}}=10^{-2,7}\text{(}\text{m}\text{o}\text{l}\text{)}\to {\text{m}}_{\text{N}\text{a}\text{O}\text{H}}=40.10^{-2,7}=0,0798\approx 0,08\text{(}\text{g}\text{)}

Đáp án: 0,3

[H⁺] = 10^-pH = 10^–4,5 = 0,3162.10^–4 → a ≈ 0,3.