Question

STEELE > Activities and Due Dates > HW 14A Homeowork #1: Buffers mt Score: 27.1% Resources [ Give Up? Hint of 11 > Calculate the change in pH when 8.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M NH, (aq) and 0.100 M in NH Cl(aq). Consult the table of ionization constants as needed. ApH = 1 Calculate the change in pH when 8.00 mL of 0.100 M NaOH is added to the original buffer solution ApH = Question Source MRG General Chemistry | Publisher Unive MacBook Air

Answer 1

Volume of Buffer solution = 100 ml [NHYOH] = 0.1 [nkyl] = 0,1M pKbCNHOH) = 4.75 for a basic buffer solution. por = pk6 + log [salt pOH = 4.75 + log of pot=4.75 pH = 14 -poh , e. pH = 9.25 Now, on adding 8 ml of 0.1 m ril. millimales of red = Molarity & Volume Cul) = 0,1 x 8 - 0,8 mmoles of Hel. monoles of Why OH = wonder of Why I a lo masles. Alinity Reagent I NH₃OH Cop.) + (Hel Cope) NHẠl (op.) + Ho t-te de lo s tot mmplo forward to tato ll. -0,8 -9.2 - 10 +0,8 = 10.9 Total volume = volume of buffer of Volume of Hel called New pote = 100 ml + 8 ul. = 108 ul. New concentator , [NHY OH = 9.2 = 0.085M 103 INHyd ] = 10,8 OM 108 New por = 4.75 + log on s pH=9. 18 por'= 4.82 0.035 APH 9.18 - 9.25 0.00 TDPH = -0.07 2. modes of Nowh = 0.8 mendes Total volume of solution = 100+8 = log ul. NHYOH (af) + shet - Nhyll ff Naonou step. 10 - Rockwooldlo 0.8 0.8 10-0,8 - 9.2 . [WHY OH] = 100 = 0,1 M , [Nhyll] = 92 = 0.085" M pot = 4.75 + log 0,085 APH = pufinal - Hinitial 9.32 - 9.25 DPH 0,077 CScarlned with CamScanner to Rachood ut lo 1. POH = 4.68 C PH = 9.32