Question

calculating the ph of a weak acid titrated with a strong base

An analytical chemist is titrating 228.6 ml. of a 0.3700 M solution of nitrous acid (HNO,) with a 0.9900 M solution of KOH. The pk of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 93.21 ml of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to 2 decimal places. | oh 1 x S ?

Answer 1

Titration of Weak acid with strong base molarity of HNO2 = 0.370 M volume of HNO2 = 228.6 mL = 0.2286 L molarity of KOH = 0.990 M volume of KOH = 93.21 mL = 0.09321 L after addition of KOH Calculating initial number of moles: moles = Molarity Volume number of moles of HNO2 = 0.370 x 0.2286 = 0.0846 mol number of moles of KOH = 0.990 X 0.09321 = 0.0923 mol total volume = 0.2286 + 0.09321 = 0.3218 L HNO2 + KOH → KNO2 + H2O initial moles 0.0846 0.0923 change in moles 0.0846 0.0846 final moles 0.0000 0.0077 left moles of KOH = 0.0077 mol 2.09321 = 0.0923 mo Calculating Concentration : concentration = moles / Volume concentration of KOH = 0.0077 / 0.3218 = 0.0239 M concentration of KOH = Concentration of OH Concentration of OH = 0.0239 M Calculating pOH: [OH] = 0.0239 M pOH = -log[OH] pOH = -log [ 0.02391 pOH = 1.62 Calculating pH: pH = 14 - POH pH = 14 -1.62 pH = 12.38