Question

f) pH after adding 25.0 mL NaOH (Equivalence point). At this point 0.00250" ist 00250 mol of OH have been added, and therefore all the acid (HCO2H) has been converted into its conjugale the table below. 25.0 mL x L/1000 mL = 0.025 L 0.025 L x 0.1 mol/L = 0.0025 mol of NaOH added n converted into its conjugate base (HCO2-). Check HCO2H 0.0025 mol HCO2- 0.000 mol Before addition addition After addition OH 0.000 mol 0.0025 mol 0.000 mol 0.00 mol 0.0025 mol At this point, the solution is no longer a buffer (There is no more acid). The solution start behaving as a weak base. Calculate the initial Molar concentration for (HCO2-] (at equivalence point). Then, calculate the pH as an quilibrium problem (Write dissociation equation, ICE table, you need Kb). OH- (aq) + HCO2H (aq) = HCO2- (aq) + H20 (1) Initial Change Equilibrium

The original given concentration was 25.0mL of 0.100 M HCO2H (formic acid Ka= 1.8x10^-4) with 0.100 M of NaOH.

Answer 1

The initial [HCO₂^-] = 0 M

The change in [HCO₂^-] = 0.0025 mol/{(25+25)/1000 L} = 0.05 M

The equilibrium [HCO₂^-] = 0 + 0.05 = 0.05 M

pH = 7 + 1/2 (pKa - Log[HCO₂^-]) = 7 + 1/2 {-Log(1.8*10^-4) - Log(0.05)} = 9.52