Respuesta :
Answer:
a. 122.6 mL / pH = 4.03
b. 193.6 mL / pH = 5.73
Explanation:
In the equivalence point we know: mmoles acid = mmoles base- And the pH in a titration between a weak base and a strong base, is acid, at the equivalence point. For the volume, we can replace the equation with the data given.
a. 0.125M . volume of acid = 65.5 mL . 0.234M
Volume of acid = (65.5 . 0.234) / 0.125 = 122.6 mL
Total volume at the equivalence point = 188.1 mL
b. 0.125M . volume of acid = 21.8 mL . 1.11 M
Volume of acid = (21.8 . 1.11) / 0.125 = 193.6 mL
Let's calculate the pH. In the equilavence point we have a neutralization reaction.
a. NH₃ + HCl → NH₄Cl
All the mmoles of protons (65.5 mL . 0.234M) react to ammonia, to obtain ammonium.
New concentration is: 15.32 mmoles / 188.1 mL = 0.0814 M
This is the [NH₄⁺] to determine the pH in the acid base equilibrium.
NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺ Ka
Expression for Ka = [NH₃] . [H₃O⁺] / [NH₄⁺]
5.6×10⁻¹⁰ = x² / (15.32 - x)
(We can avoid the quadratic equation 'cause Ka is so small)
√(5.6×10⁻¹⁰ . 15.32) = x → [H₃O⁺] = 9.26×10⁻⁵
pH = - log [H₃O⁺] → 4.03
b. CH₃NH₂ + HCl → CH₃NH₃Cl
All the mmoles of protons (21.8 mL . 1.11M) react to methylamine, to obtain methylammonium.
New concentration is: 24.2 mmoles / 193.6 mL = 0.125 M
This is the [CH₃NH₃⁺] to determine the pH in the acid base equilibrium.
CH₃NH₃⁺ + H₂O ⇄ CH₃NH₂ + H₃O⁺ Ka
Expression for Ka = [CH₃NH₂] . [H₃O⁺] / [CH₃NH₃⁺]
2.7×10⁻¹¹ = x² / (0.125 - x)
(We can avoid the quadratic equation 'cause Ka is so small)
√(2.7×10⁻¹¹ . 0.125) = x → [H₃O⁺] = 1.84×10⁻⁶
pH = - log [H₃O⁺] → 5.73
The pH of the equivalence point of 0.125M HCl with NH₃ is 4.03 and with CH₃NH₂ is 5.73.
What is equivalence point?
In the acid - base titration, at the equivalence point equal moles of acid as well as of base are present.
First we calculate the volume of HCl by using the below equation as:
- M₁V₁ = M₂V₂, where
M₁ = molarity of NH₃ = 0.234 M
V₁ = volume of NH₃ = 65.5 mL
M₂ = molarity of HCl = 0.125 M
V₂ = volume of HCl = ?
V₂ = (65.5 . 0.234) / 0.125 = 122.6 mL
Total volume at the equivalence point = 188.1 mL
Chemical reacion will be written as:
NH₃ + HCl → NH₄Cl
All moles of protons react with moles of ammonia, so new moles (n) will be calculated as:
n = molarity × volume
n = 65.5 mL × 0.234M = 15.32 moles
Now, concentration in terms of molarity will be 15.32 mmoles / 188.1 mL = 0.0814 M
pH in the acid-base reaction due to [NH₄⁺], as:
NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺
Value of Ka for this reaction is = 5.6×10⁻¹⁰
According to the ICE table, Ka equation will be written as-
5.6×10⁻¹⁰ = x² / (15.32 - x)
x = [H₃O⁺] = 9.26 × 10⁻⁵
pH = - log [H₃O⁺] = 4.03
- Again we calculate the volume of HCl with the CH₃NH₂ as:
0.125M . volume of acid = 21.8 mL . 1.11 M
Volume of acid = (21.8 . 1.11) / 0.125 = 193.6 mL
Total volume at the equivalence point = 215.4 mL
Chemical reacion will be written as:
CH₃NH₂ + HCl → CH₃NH₃Cl
All moles of protons react with moles of ammonia, so new moles (n) will be calculated as:
n = 21.8 mL . 1.11M = 24.2 moles
Now, concentration in terms of molarity will be 24.2 moles / 215.4 mL = 0.11M
pH in the acid-base reaction due to [CH₃NH₃⁺], as:
CH₃NH₃⁺ + H₂O ⇄ CH₃NH₂ + H₃O⁺
Value of Ka = 2.7×10⁻¹¹
According to the ICE table, Ka equation will be written as-
2.7×10⁻¹¹ = x² / (0.11 - x)
x = [H₃O⁺] = 5.4×10⁻⁶
pH = - log [H₃O⁺] = 5.73
Hence the value of pH will be 4.03 and 5.73 respectively.
To know more about equivalence point, visit the below link:
https://brainly.com/question/24584140
