Question

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

An analytical chemist is titrating 52.4 ml of a 1.200 M solution of butanoic acid (HC,H,CO) with a 0.9500 M solution of KOH. The pK, of butanoic acid is 4.82. Calculate the pH of the acid solution after the chemist has added 20.7 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. DH- X ? Explanation) Check

Answer 1

M mL = 0.0524 L M ml = 0.0207 L Titration of Weak acid with strong base molarity of C3H7COOH = 1.200 volume of C3H7COOH = 52.4 molarity of KOH = 0.950 volume of KOH = 20.7 pka = 4.82 after addition of KOH C3H7COOH + KOH Weak acid Calculating initial number of moles: moles = Molarity Volume initial moles of C3H2COOH = 1.200 initial moles of KOH = 0.950 + C3H7COOK + H2O salt X X 0.0524 0.0207 = 0.0629 mol = 0.0197 mol 0 C3H7COOH + KOH + C3H7COOK + H2O initial moles 0.0629 0.0197 change in moles 0.0197 0.0197 0.0197 final moles 0.0432 0.0000 0.0197 left moles of C3H2COOH = 0.0432 mol formed moles of C3H7COOK = 0.0197 mol According to Henderson-Hasselbalch equation pH = pka + log[salt/acid] pH = 4.82 +log [ 0.0197 / 0.0432 ] pH = 4.48