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

there is none because earth is stationary and flat.

warm front forms when a moist, warm air mass slides up and over a cold air mass. as the warm air mass rises, it condenses into a broad area of clouds. a warm front brings gentle rain or light snow, followed by warmer, milder weather.

hope i !