Respuesta :
Answer:
pH → 1.83
Explanation:
Let's write the lactic acid as HLac, because it is a monoprotic weak acid.
HLac + H₂O ⇄ H₃O⁺ + Lac⁻ Ka
Initially we have 0.25 moles, so x amount has reacted. At the end, when the equilibrium is finished, we may have (0.25 - x) moles of acid, x moles of protons have been released and x moles of lactate were formed.
In order to find x, we use the acid constant, Ka. The expression for Ka is:
Ka = [H₃O⁺] . [Lac⁻] / [HLac]
8.3×10⁻⁴ = x² / (0.25-x)
8.3×10⁻⁴ (0.25 - x) = x²
8.3×10⁻⁴ . 0.25 - 8.3×10⁻⁴x - x² → this is a quadractic function
a = -1, b = 8.3×10⁻⁴, c = 2.07×10⁻⁴
We solve: (-b - √(b² - 4ac) / (2a)
(-8.3×10⁻⁴ - (√ ((8.3×10⁻⁴)² - 4 (-1) (2.07×10⁻⁴) ) / (2 . -1) = 0.0148
[H₃O⁺] = 0.0148 M
- log [H₃O⁺] = pH → - log 0.0148 = 1.83
The pH is defined as the power of Hydrogen. The pH of a 0.25 M solution of lactic acid is 1.83.
[tex]\bold { CH_3COOH \rightarrow CH_3 COO + H^+\ \ \ \ \ \ \ \ \ \ \ \ (Ka = 8.3 x 10 -4 ).}[/tex]
Initial concentration of lactic acid is 0.25 M, So, [tex]x[/tex]. At the end of equilibrium [tex]x[/tex] amount of lactic acid dissociates.
So, concentration of the Hydrogen ion in the (0.25 - [tex]x[/tex])
[tex]x[/tex] can be found out by using the dissociation constant formula,
[tex]\bold {Ka = \dfrac {[H^+][CH_3COO^-]}{[Ch_3COOH]}}\\[/tex]
[tex]\bold {8.3x10^-^4= \dfrac {[x][x]}{[2.5- x]}}\\\\\bold {8.3x10^-^4= \dfrac {[x^2]}{[2.5- x]}}\\\\\bold {x^2= 8.3x10^-^4\times [2.5- x]}}\\\\\bold {8.3x10^-^4 \times 0.25 - 8.3x10^-^4 x - x^2 = 0}[/tex]
The equation above is a quadratic equation, solving it we get,
The concentration of protons = 0.0148 M
Put this into the pH formula,
[tex]\bold {pH = -log [H^+]}[/tex]
[tex]\bold {pH = -log [ 0.0148]}\\\\\bold {pH = 1.83}[/tex]
Therefore, the pH of a 0.25 M solution of lactic acid is 1.83.
To know more about pH,
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