Answer:
see explanation below
Step-by-step explanation:
In this case, we need to write the overall reaction:
HC₃H₅O₃ + H₂O <-------> C₃H₅O₃⁻ + H₃O⁺
The lactic acid is a weak acid, so, when it dissociates in it's ions, part of the acid is dissociated. This depends of it's Ka to know which quantity was dissociated.
To calculate Ka, let's write an ICE chart first:
HC₃H₅O₃ + H₂O <-------> C₃H₅O₃⁻ + H₃O⁺ Ka = ?
i) 0.045 0 0
c) -y +y +y
e) 0.045 - y y y
Writting the Ka expression we have:
Ka = [C₃H₅O₃⁻] [H₃O⁺] / [HC₃H₅O₃]
Now, to calculate Ka we need the values of [C₃H₅O₃⁻] and [H₃O⁺] in equilibrium. Fortunately, we have the value of the pH, which is 2.63 and with this we can get the value of [H₃O⁺] and then, the value of y. With that value, we replace it in the Ka expression to calculate Ka:
[H₃O⁺] = 10^(-pH)
[H₃O⁺] = 10^(-2.63)
[H₃O⁺] = [C₃H₅O₃⁻] = x = 2.34x10⁻³ M
Now, let's replace this value in the Ka expression:
Ka = (2.34x10⁻³)² / (0.045 - 2.34x10⁻³)
Ka = 1.28x10⁻⁴
b) Now, let's calculate the pH with the obtained value of Ka. We will use the same expression of Ka so:
1.28x10⁻⁴ = y² / (0.045-y)
1.28x10⁻⁴ (0.045 - y) = y²
5.76x10⁻⁶ - 1.28*10⁻⁴y = y²
y² + 1.28x10⁻⁴y - 5.76x10⁻⁶ = 0
From here, we'll use the quadratic equation general formula, for solving y:
y = -1.28x10⁻⁴ ±√(1.28x10⁻⁴)² + 4 * 1 * 5.76x10⁻⁶ / 2
y = -1.28x10⁻⁴ ±√2.31x10⁻⁵ / 2
y = -1.28x10⁻⁴ ± 4.8x10⁻³ / 2
y₁ = 2.34x10⁻³ M
y₂ = -2.464x10⁻³ M
Now, as y₁ is positive this is the value that we will take.
This value would be the [H₃O⁺] in equilibrium.
The value of pH would be:
pH = -log[H₃O⁺]
pH = -log(2.34x10⁻³)
pH = 2.631
According to this value of pH we can actually expect an inprovement in the accuracy, basically because we obtain a value with more significant figures, and this are relationed with accuracy.