Question

2. Consider an aqueous solution comprised of 0.200 M NH3 and 0.150 M NHACI. The pKb of NH3 is 4.75 and the volume of the solution is 1.00 L. a) (2 marks) What is the pH of this solution? Show your work. b) (1 mark) What is the chemical reaction (balanced chemical equation) that occurs when NaOH solution is added to the original solution? c) (3 marks) How many moles of NaOH(s) could this solution react with before its pH rises to 10.00? Show your work.

Answer 1

12 d2 Then Henderson - Hauselbach equation can be written as: PH 2 pka + log [conjugate base] [acid ] or pot apkb & Log (conjugate acid] [base] . Given NH3 0.200 M pkp (NH₃) 24.75 NH y C 0. 150 M vol. 2 1.0L NH₃ + H₂O - NHi+ OH- Nhy C – NHitte- Using Henderson Havelbach equation, pota 4.75 + Log (NH4t I POH 2 4.75 + log ( 0.150) poH 4.625 pH 14-POH 27 (pH=9.875 | 4.75 POH 2 0.150 0.200

(6) When NaOH solution is added to original solution, the OH- ions combine with NHy ions to given NH₃ and H₂O. This decreases the pH. and so buffer is able to maintain the pH. NHut + oH - NH3 + H₂O LC) Now we need to calculate the no. of moles of Naoh required to reach PH 210. i pOH 2 4. lets say we added a moles of NaOH. Then & moles od on will combine with x moles of NHut to give water and ammonia. NHYT + ou - NH₃ +H2O 0.150 x But only a moles of Nhut can combine Moles of NHut left a 0.150-4. Using the Henderson - Horsebach equation again, pot - pkt (conjugate acid] [base] 4e 4.75 + log (0.150 [ 0.200 J antilog (0.75 a 0.150 - 0 200 In a 00114 Moles of Nach requested = 0.114)