Answer:
pH = 3.57 after adding 10ml of 0.125M NaOH
Explanation:
NaOH + HCHO₂ => NaCHO₂ + H₂O
=> 10ml(0.125M NaOH) + 30ml(0.030M HCHO₂)
=> 0.01(0.125)mol NaOH + 0.036(0.030)mol HCHO₂ ***(see reference note at end of problem)
=> 0.00125mol NaOH + 0.00315mol HCHO₂
=> (0.00315mol - 0.00125mol) HCHO₂ = 0.0019mol HCHO₂ in excess
+ 0.00125mol NaCHO₂ formed.
HCHO₂ ⇄ H⁺ + CHO₂⁻
C(i): 0.0019mol/0.040L 0.00M 0.00125mol/0.40L
= 0.0475M = 0.0313M
ΔC: -x +x +x
C(eq): 0.0475 - x x 0.0313 +x
Ka = [H⁺][CHO₂⁻]/[HCHO₂] = (X)(0.0313 + x)/(0.0475) = 1.8 X 10⁻⁴
=> x² + 0.03168x + (-8.55 x 10⁻⁶) = 0
Solve with quadratic equation ...
x = 2.68 x 10⁻⁴M = [H⁺] at equilibrium
pH = - log[H⁺] = -log(2.68 x 10⁻⁴) = -(-3.57) = 3.57
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*** remember, => moles = Molarity x Volume in Liters
