Question

Calculate the change in pH when 3.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH, (aq) and 0.100 M in NHCl(aq). Consult the table of ionization constants as needed. ApH = Calculate the change in pH when 3.00 mL of 0.100 M NaOH is added to the original buffer solution. ApH =

Answer 1

13 NHAU 0.100M 0.100 M of buller = loom Volume ImL = 0.0012 100ml = 0.12 Initial mol of NH₃ = Molarity & volume = Oil mol Xonly = 0.01 mol 1 -0.01 mol - Initial mol of NHAU = 0.1 mol xo.14 When He added to butler solution, itu mol og Ha added = 0.003 Y x 0.1 mol = 0.0003 mol to butter solution, it wire react with weak base present in butter and amount of weak base decreases and that conjugate and increases. After addition mol og N43 left = 0.01 - 0.0003 = 0.0097 mol moly N44c6 formed: 0.01 +0.003 = 0.0103 mo) Total volume - 01L+ 0.0032 = 0.1034 [M43 ) = 0.0097901 0.103 2 0.103 Hasseltach equation (NH4U)= 0.0103 mo! pH of butter Calculated using Henderson pu= bka + log [unjugate fase] Сеяк а ј = 9.26+ 609 [0.0103 ] Ortos [0.0097) Oiles 14 = 9.99) buy solution = 9.26

M43 Nhace 0.IM O.IM solution using Hendeson Hasselbach equation brey Nhace= 9.26 Calculating initial pH of buffer pu= pka + log [nnace] CNH] = 9.86+ los content bu = 926 Initial pu= 9.26 Du 7 solution after addition g 466 = 9.29 Change in pu= l final pn- initial pul 9.29 - 1.26 = 0.03 Change in pu= 0.03

moly naoh added = 0.003 mol ① mol of Noor moly Naon added = 0.0 by OX 0.003% = 0.0003 mol When Naon addel, it will react with Conjugate and present in butter and molo conjugate acid decreases and weak acis increme. Afler addism moly N43 Semned = 0:01+0.0003 = 0.0103 moj molg N1444 left = 0.01 – 0.0003 = 0.0097 0.0097 0103 [N437 = 0.0103 mo wnae) = 103 L bu= 9.26+ log c 0.0097) 0.105 = 9.23 0.0003 0.45 Change in pn=19.26 -9.231 = 0.03 unit