An analytical chemist is titrating 59.1 mL of a 0.6700 M solution of ammonia (NH,) with a 0.4500 M solution of HIO3. The p K, of ammonia is 4.74. Calculate the pH of the base solution after the chemist has added 16.2 mL of the HIO, solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HIO, solution added. Round your answer to 2 decimal places. pH =

Given : Concentration of NH3 = 0.6700 M Volume of NH3 solution = 59.1 mL = 0.0591 L (since 1 L = 1000 mL) Concentration of HIO3 = 0.4500 M Volume of HIO3 solution = 16.2 mL = 0.0162 L And pKb of NH3 = 4.74 Since moles = concentration X volume of solution in L => Moles of NH3 = 0.6700 X 0.0591 = 0.039597 mol And moles of HIO3 = 0.4500 X 0.0162 = 0.00729 molSince HIO3 is a strong acid. Hence it will react completely with NH3 as per the reaction, => NH3 + HIO3 -------> NH4IO3 From the above reaction we can see that 1 mole of HIO3 is reacting with 1 mole of NH3 to produce 1 mole of NH4IO3 Hence moles of NH3 reacted = moles of HIO3 taken = 0.00729 mol And moles of NH4IO3 formed = moles of HIO3 taken = 0.00729 mol Hence moles of NH3 remaining in the solution = taken - reacted = 0.039597 - 0.00729 = 0.032307 mol Since the solution has both, weak base NH3 and its conjugate acid salt NH4IO3. Hence the solution is a buffer solution.Since the pOH of base buffer is given by Henderson-Hasselbalch equation as where pKb = 4.74 [salt] = concentration of conjugate acid of base i.e NH4IO3 [base] = concentration of base i.e NH3 Since the final volume of solution will be same for both salt and base. Hence the ratio of concentration inside log will be same as ratio of moles. Hence substituting the values we get, => pOH = 4.74 + log0.007290.032307 = 4.09 approx. => pH = 14 - pOH = 14 - 4.09 = 9.91 approx.