Calculate the pH after 10.0 mL of 0.125 M NaOH is added to 30.0 mL of 0.105 M HCHO2. Ka (HCHO2) = 1.8 × 10-4

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

*** remember, => moles = Molarity x Volume in Liters

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation: