Question

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10?5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 28.0 mL of HNO3. Express your answer numerically.

Answer 1

Given that NH,J 0.200 M Volume of NH3=75.0mL HNO,] 0.500 M Volume of HNO3 28.0 mL K for NH 1.8x105 Calculate the value of pK,, pK -log K -log (1.8x10 =4.74 Calculate the number of moles of ammonia before the reaction Moles of NH Conentrati on of NH3xVolume of NH3 in L X mol x 75.0 pL -0.200 1000mL 0.015 mol Calculate the number of moles of HNO, before the reaction, Moles of HNO3= Conentration of HNOxVolume of HNO mol x28.0 mL*1000 L 0.500 0.014mol Consider the reaction NH (aq) HNO, (aq) NH2 (a) NO (aq) > + Initial moles (in mol) 0.015 0.014 0 Change in moles 0.014 -0.014 -0.014 At equilibrium 0.001 0.014

completely consumed in the reaction. This At equilibrium, the number of moles of HNO are means that the solution contains the number of moles of weak base, ammonia and the number of moles of salt, ammonium nitrate (NH,NO) Calculate the concentration of weak base at equilibrium Moles of reamaining NH Concentration of remaining NH Volume of solution in L 1000 L 0.001mol (75.0+28.0) mL 1L 0.0097mol L1 0.0097 M Calculate the concentration of salt (NH^NO) Moles of NH NO Concentration of NH4NO; Voume of solution in L 1000 pL 0.014mol (75.0+28.0) mL 1L -0.1359mol L -0.1359M Weak acid and its conjugate base forms a buffer solution, therefore, to know the pH of the solution, use the Henderson-Hasselbalch equation [Conjugate acid] [Weak base] рон-рк, +1og [сн,соONa] [сH, соон] рH-4.70+1og 0.1359M РОН-4.74 +1og 0.0097 M pOH 4.74 46) -5.89 Calculate the pH for the solution by using following relation

PH +рОН %314 рH-14-рОН =14-5.89 -8.11 Hence, pH for the solution is 8.11|