Question

Formic acid has a Ka of 1.77*10^-4. To 55.0 mL of 0.25 M solution 75.0 of 0.12 M NaOH is added . What is the resulting pH.

Answer 1

The first step is to determine the moles of formic acid and NaOH that react. moles of formic acid = (0.25 mol/L) x (0.055 L) = 0.01375 mol moles of NaOH = (0.12 mol/L) x (0.075 L) = 0.009 mol Since NaOH is a strong base, it will react completely with formic acid to form sodium formate and water: HCOOH + NaOH → HCOONa + H2O The limiting reagent in this case is NaOH, so all of it will be consumed in the reaction. The amount of excess formic acid that remains can be calculated: moles of HCOOH remaining = moles of HCOOH initial - moles of NaOH used moles of HCOOH