Dd CH3COONa 0,10M (Kb của CH3COO- = 5,71.10-10 )Tìm pH ?
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- Đáp án D
- Do CH3COOH là chất điện li yếu nên trong nước chỉ phân li một phần
CH3COOH ⇌ H+ + CH3COO-
Vì vậy [H+] < [CH3COO-]= 0,1M
2CH3COOH+Na->CH3COONa+H2
CH3COOH+Na2CO3->CH2COONa+h2O+CO2
2CH3COOH+CaCO3->(CH3COO)2Ca+H2O+CO2
2CH3COOH+Mg->(CH3COO)2Mg+H2
Bài 4: Hoàn thành các PTHH sau:
a, Na + CH3COOH → CH3COONa + \(\dfrac{1}{2}\)H2
b, Na2CO3 + 2CH3COOH → 2CH3COONa + H2O + CO2
c, 2CH3COOH + CaCO3 → (CH3COO)2Ca + H2O + CO2↑
d, 2CH3COOH + Mg → (CH3COO)2Mg + H2↑
a) \(2Na+2CH_3COOH\rightarrow2CH_3COONa+H_2\)
b) \(Na_2CO_3+2CH_3COOH\rightarrow2CH_3COONa+CO_2+H_2O\)
c) \(2CH_3COOH+CaCO_3\rightarrow\left(CH_3COO\right)_2Ca+H_2O+CO_2\)
d) \(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
\(a) 2Na+2CH_3COOH→2CH_3COONa+H_2 \)
\(b) Na_2CO_3+2CH_3COOH→2CH_3COONa+CO_2+H_2O \)
\(c) 2CH_3COOH+CaCO_3→(CH_3COO)_2Ca+H_2O+CO_2↑ \)
\(d) 2CH_3COOH+Mg→(CH_3COO)_2Mg+H_2↑\)
$C_2H_4 + H_2O \xrightarrow{t^o,H^+} C_2H_5OH$
$C_2H_5OH + CH_3COOH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
$CH_3COOC_2H_5 + H_2O \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOH + C_2H_5OH$
$2CH_3COOH + Mg \to (CH_3COO)_2Mg + H_2$
$(CH_3COO)_2Mg + Ca(OH)_2 \to (CH_3COO)_2Ca + Mg(OH)_2$
$(CH_3COO)_2Ca + K_2CO_3 \to 2CH_3COOK + CaCO_3$
$C_6H_{12}O_6 \xrightarrow{t^o,men\ rượu} 2CO_2 + 2C_2H_5OH$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
$C_2H_5OH + CH_3COOH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
$CH_3COOC_2H_5 + H_2O \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOH + C_2H_5OH$
$2CH_3COOH + CuO \to (CH_3COO)_2Cu + H_2O$
$(CH_3COO)_2Cu + NaOH \to 2CH_3COONa + Cu(OH)_2$
\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)
\(2Na+2C_2H_5OH\rightarrow2C_2H_5ONa+H_2\)
\(Fe_2O_3+6CH_3COOH\rightarrow2\left(CH_3COO\right)_3Fe+3H_2O\)
\(2CH_3COOH+CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)
\(CH_3COOH+CH_3OH\underrightarrow{H_2SO_{4\left(đ\right)},t^o}CH_3COOCH_3+H_2O\)
\(6CH_3COOH+2Al\rightarrow2\left(CH_3COO\right)_3Al+3H_2\)
Chọn đáp án D
Đây là 2 dung dịch có môi trường bazo.NaOH có tính bazo mạnh hơn nên x >y.
CH3COONa ---> CH3COO(-) + Na(+)
0.1_____________0.1
CH3COO(-) + H2O ---> CH3COOH + OH(-)
Bđ 0.1 0 0
Pư x x x
Cb 0,1-x x x
=> x2/(0,1-x) = 5,71.10-10
<=>x = 7,556.10-6 => pOH = 5,121 => pH = 8.878.