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 =
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 <em>pH = - log [H⁺]</em>
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 = = 2<em>,5x10⁻³ mol</em>
KOH = = 1<em>,0x10⁻³ mol</em>
CH₃COOK = 0.
In equilibrium:
CH₃COOH = 2,5x10⁻³ mol - 1,0x10⁻³ mol =<em> </em>1<em>,5x10⁻³ mol</em>
KOH = 0 mol
CH₃COOK = 1<em>,0x10⁻³ mol</em>
Now, you can use Henderson–Hasselbalch equation:
pH = 4,76 + log
pH = 4,58
c) With adition of 10,0 mL of 0,200M KOH solution the initial moles are:
CH₃COOH = = 2<em>,5x10⁻³ mol</em>
KOH = = 2<em>,0x10⁻³ mol</em>
CH₃COOK = 0.
In equilibrium:
CH₃COOH = 2,5x10⁻³ mol - 2,0x10⁻³ mol =<em> 0,5x10⁻³ mol</em>
KOH = 0 mol
CH₃COOK = 2<em>,0x10⁻³ mol</em>
Now, you can use Henderson–Hasselbalch equation:
pH = 4,76 + log
pH = 5,36
d) With adition of 12,5 mL of 0,200M KOH solution the initial moles are:
CH₃COOH = = 2<em>,5x10⁻³ mol</em>
KOH = = 2<em>,5x10⁻³ mol</em>
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 =
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 <em>pOH = - log [OH⁻]; pH = 14 - pOH</em>
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 × ÷ 0,040 L = 0,0125 = [OH⁻]
<em>pOH = - log [OH⁻]; pH = 14 - pOH</em>
pOH = 1,90
pH = 12,10
I hope it helps!