nyons-CHEM 202-Spring18-FLYNN Activities and Due Dates HW 14 Part 2 0 4/9/2018 1

Question

nyons-CHEM 202-Spring18-FLYNN Activities and Due Dates HW 14 Part 2 0 4/9/2018 1155 PM 6.8/10 /8/2018 12:35 AM Print CalculatorPerlodic Table 9 Gradeb Question 13 of 20 oChomistry 4th EditionUniversity Science Books Map presented by Saping Leaming Calculate the change in pH when 6.00 mL of 0.100 M HCI(ag) is added to 100.0 mL of a buffer solution that is 0.100 M in NHs(aq) and 0.100 M in NH4CI(aq). A list of ionization constants can be found here Number Calculate the change in pH when 6.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution. Number Exit ) Check Answer ONex! ) Previous Cive, up & View Sokne

Explanation / Answer

1) We find pH initially when only NH3 and NH4Cl were present ....as it is a basic buffer mixture ...
so pOH = pKb + log [NH4Cl]/[NH3]
pKb for NH3 = 4.745 , So pOH = 4.745 + log 0.1/0.1

pOH = 4.745 + log 1
pOH = 4.745 + 0 = 4.745
and pH = 14 - pOH = 14 - 4.745 = 9.255
initial no.of moles of NH3 = molarity X volume in litres = 0.1 X 100/1000 = 0.01
no.of moles of NH4Cl = 0.1 X 100/1000 = 0.01
no.of moles of HCl = 0.1 X 6/1000 = 6 X 10^-4
now when you add HCl ...following reaction will take place
NH3 + HCl -------> NH4Cl
so more NH4Cl will be formed
as HCl is present in lesser amount ( 6 X 10^-4) than NH3 ( 0.01) so HCl is the limiting reagent
no.of moles of NH4Cl formed = 6 X 10^-4
new no.of moles of NH4Cl = 0.01 + 6 X 10^-4 = 0.0106
new no.of moles of NH3 = 0.01 - 6 X 10^-4 = 0.0094
total volume = 6 + 100 = 106 ml = 0.106 L
new [NH3] = 0.0094/0.106 = 0.0887 M
[NH4Cl] = 0.0106/0.106 = 0.1 M
new pOH = 4.745 + log 0.1/0.08876
pOH = 4.745 + log 1.128
pOH = 4.745 + 0.0522 = 4.7972
new pH = 14 - 4.7972 = 9.2028
change in pH = 9.2028 - 9.255 = - 0.0522

2) no.of moles of NaOH = 0.1 X 6/1000 = 6 X 10^-4
this time this reaction will take place :-
NH4Cl + NaOH -------> NH3 + NaCl + H2O
so this time NH4Cl is consumed and NH3 is formed
new no.of moles of NH4Cl = 0.01 - 6 X 10^-4 = 0.0094
and no.of moles of NH3 = 0.01 + 6 X 10^-4 = 0.0106
new [NH4Cl] = 0.0094/0.106 = 0.0887 M
[NH3] =0.0106/0.106 = 0.1 M
pOH = 4.745 + log 0.0887/0.1
pOH = 4.745 + log 0.887
pOH = 4.745 - 0.0522 = 4.693
so new pH = 14 - 4.693 = 9.307
change = 9.307 - 9.255 = 0.0522.

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