Respuesta :
Answer:
a) pH = 2,88
b) pH = 4,58
c) pH = 5,36
d) pH = 8,79
e) pH = 12,10
Explanation:
In a titration of a strong base (KOH) with a weak acid (CH₃COOH) the reaction is:
CH₃COOH + KOH → CH₃COOK + H₂O
a) Here you have just CH₃COOH, thus:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ where ka =1,74x10⁻⁵ and pka = 4,76
When this reaction is in equilibrium:
[CH₃COOH] = 0,100 -x
[CH₃COO⁻] = x
[H⁺] = x
Thus, equilibrium equation is:
1,74x10⁻⁵ = [tex]\frac{[x][x] }{[0,100-x]}[/tex]
The equation you will obtain is:
x² + 1,74x10⁻⁵x - 1,74x10⁻⁶ = 0
Solving:
x = -0,0013278193 ⇒ No physical sense. There are not negative concentrations
x = 0,0013104193
As x = [H⁺] and pH = - log [H⁺]
pH = 2,88
b) Here, it is possible to use:
CH₃COOH + KOH → CH₃COOK + H₂O
With adition of 5,0 mL of 0,200M KOH solution the initial moles are:
CH₃COOH = [tex]0,025 L.\frac{0,100 mol}{L} =[/tex] = 2,5x10⁻³ mol
KOH = [tex]0,005 L.\frac{0,200 mol}{L} =[/tex] = 1,0x10⁻³ mol
CH₃COOK = 0.
In equilibrium:
CH₃COOH = 2,5x10⁻³ mol - 1,0x10⁻³ mol = 1,5x10⁻³ mol
KOH = 0 mol
CH₃COOK = 1,0x10⁻³ mol
Now, you can use Henderson–Hasselbalch equation:
pH = 4,76 + log [tex]\frac{1,0x10^{-3} }{1,5x10^{-3} }[/tex]
pH = 4,58
c) With adition of 10,0 mL of 0,200M KOH solution the initial moles are:
CH₃COOH = [tex]0,025 L.\frac{0,100 mol}{L} =[/tex] = 2,5x10⁻³ mol
KOH = [tex]0,010 L.\frac{0,200 mol}{L} =[/tex] = 2,0x10⁻³ mol
CH₃COOK = 0.
In equilibrium:
CH₃COOH = 2,5x10⁻³ mol - 2,0x10⁻³ mol = 0,5x10⁻³ mol
KOH = 0 mol
CH₃COOK = 2,0x10⁻³ mol
Now, you can use Henderson–Hasselbalch equation:
pH = 4,76 + log [tex]\frac{2,0x10^{-3} }{0,5x10^{-3} }[/tex]
pH = 5,36
d) With adition of 12,5 mL of 0,200M KOH solution the initial moles are:
CH₃COOH = [tex]0,025 L.\frac{0,100 mol}{L} =[/tex] = 2,5x10⁻³ mol
KOH = [tex]0,0125 L.\frac{0,200 mol}{L} =[/tex] = 2,5x10⁻³ mol
CH₃COOK = 0.
Here we have the equivalence point of the titration, thus, the equilibrium is:
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ kb = kw/ka where kw is equilibrium constant of water = 1,0x10⁻¹⁴; kb = 5,75x10⁻¹⁰
Concentrations is equilibrium are:
[CH₃COOH] = x
[CH₃COO⁻] = 0,06667-x
[OH⁻] = x
Thus, equilibrium equation is:
5,75x10⁻¹⁰ = [tex]\frac{[x][x] }{[0,06667-x]}[/tex]
The equation you will obtain is:
x² + 5,75x10⁻¹⁰x - 3,83x10⁻¹¹ = 0
Solving:
x = -0.000006188987⇒ No physical sense. There are not negative concentrations
x = 0.000006188
As x = [OH⁻] and pOH = - log [OH⁻]; pH = 14 - pOH
pOH = 5,21
pH = 8,79
e) The excess volume of KOH will determine pH:
With 12,5mL is equivalence point, the excess volume is 15,0 -12,5 = 2,5 mL
2,5x10⁻³ L × [tex]\frac{0,200 mol}{1L}[/tex] ÷ 0,040 L = 0,0125 = [OH⁻]
pOH = - log [OH⁻]; pH = 14 - pOH
pOH = 1,90
pH = 12,10
I hope it helps!
Answer:
The pH of solution on addition of 0.0, 5.0, 10.0 12.5 and 15 ml of KOH will be 2.87, 4.56, 5.34, 4.74, and 12.09 respectively.
Explanation:
pH can be calculated by the evaluation of Hydronium ions in the solution.
[tex]\rm K_a[/tex] of [tex]\rm CH_3COOH[/tex] is 1.8 [tex]\rm \times10^-^5[/tex]
(a) Hydrogen ion concentration = [tex]\rm \sqrt{K_a\;\times\;CH_3COOH\;concentartion}[/tex]
Hydrogen ion concentration = [tex]\rm \sqrt{1.8\;\times\;10^-^5\;\times\;0.1}[/tex]
Hydrogen ion concentration = 1.34 [tex]\times\;10^-^3[/tex] M
pH of solution = log [Hydrogen ion concentration]
pH = log [[tex]\rm 1.34\;\times10^-^5[/tex]]
pH = 2.87
(b) On addition of 5 ml of KOH of 0.200 M, the moles of KOH added are:
moles of KOH = [tex]\rm \frac{volume\;(ml)}{1000}\;\times\;\frac{molarity}{L}[/tex]
moles of KOH = [tex]\rm \frac{5}{1000}\;\times\;\frac{0.200}{L}[/tex]
moles of KOH = [tex]\rm 1\;\times\;10^-^3[/tex] M
The initial moles of [tex]\rm CH_3COOH[/tex] are:
moles of [tex]\rm CH_3COOH[/tex] = [tex]\rm \frac{25}{1000}\;\times\;\frac{0.100}{L}[/tex]
moles of [tex]\rm CH_3COOH[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex]
At equilibrium, the concentration of [tex]\rm CH_3COOH[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex] - [tex]\rm 1\;\times\;10^-^3[/tex] mol
concentration of [tex]\rm CH_3COOH[/tex] = [tex]\rm 1.5\;\times\;10^-^3[/tex] moles.
The concentration of [tex]\rm CH_3COOK[/tex] = [tex]\rm 1\;\times\;10^-^3[/tex] moles
pH = [tex]\rm pK_a\;+\;log\;\frac{salt}{acid}[/tex]
pH = 4.76 + log [tex]\rm \frac{1\;\times\;10^-^3}{1.5\;\times\;10^-^3}[/tex]
pH = 4.58
(c) On addition of 10 ml of KOH,
moles of KOH = [tex]\rm \frac{10}{1000}\;\times\;\frac{0.200}{L}[/tex]
moles of KOH = [tex]\rm 2\;\times\;10^-^3[/tex] moles
moles of [tex]\rm CH_3COOH[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex] moles
moles of [tex]\rm CH_3COOK[/tex] = [tex]\rm 2\;\times\;10^-^3[/tex] moles
pH = 4.76 + log [tex]\rm \frac{2\;\times\;10^-^3}{0.5\;\times\;10^-^3}[/tex]
pH = 5.36
(d) On addition of 12.5 ml of KOH,
moles of KOH = [tex]\rm \frac{12.5}{1000}\;\times\;\frac{0.200}{L}[/tex]
moles of KOH = [tex]\rm 2.5\;\times\;10^-^3[/tex]
moles of [tex]\rm CH_3COOH[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex] moles
moles of [tex]\rm CH_3COOK[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex] moles
pH = 4.76 + log [tex]\rm \frac{2\;\times\;10^-^3}{0}[/tex]
pH = 4.74
(e) On addition of 15.0 ml of KOH,
The excess point is reached after the addition of 12.5 ml of KOH. After further addition of KOH, the pOH will be there.
The OH concentration on addition of KOH = [tex]\rm \frac{(\frac{15}{1000}\;\times0.200\;moles)\;-\;(\frac{25}{1000}\;\times0.100\;moles) }{\frac{25}{1000}\;+\;\frac{15}{1000} }[/tex]
= 0.0125 M
pOH = -log [OH concentration]
pOH = -log [0.0125]
pOH = 1.903
pH = 14 - pOH
pH = 14 - 1.903
pH = 12.097
The pH of solution on addtion of KOH will be :
O ml KOH = 2.87
5 ml KOH= 4.56
10 ml KOH = 5.36
12.5 ml KOH = 4.74
15 ml KOH = 12.907.
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