Question 23 of 32 Map A General Chemistr sity Science Books presented by Sapling

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Question 23 of 32 Map A General Chemistr sity Science Books presented by Sapling Lea Donald McQuarrie. Peter A Rock .Ethan Gallogly Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.210 M pyridine, C5H5N(aq) with 0.210 M HBr(aq): Number (a) before addition of any HBr Number (b) after addition of 12.5 mL of HBr Number (c) after addition of 14.0 mL of HBr Number (d) after addition of 25.0 mL of HBr Number (e) after addition of 33.0 mL of HBr L Hint O Previous ® Give Up & View Solution Check Answer Next Ext

Explanation / Answer

The reaction is

C5H5N + H2O<-->C5H5NH+ + OH- , pKb of pyridine is 8.75

before any addition of HBr

kb= [C2H5NH+] [OH-]/ [C5H5N]

Pkb for pyridine= 8.75, Kb=1.78*10-9

1.78*10-9 = x2/(0.21-x), where x=drop in concentration of pyridine to reach equilibrium, [OH-] =x

when solved using solver, x=1.93*10-5

[OH-] =1.93*10-5, pOH= 4.71, pH= 14-4.71= 9.29

b) when 12.5 ml of 0.21 HBr is added, moles of HBr= 0.21*12.5/1000=0.002625 moles

moles of pyridine =0.21*25/1000=0.00525

The reaction between pyridine and HBr is

C5H5N + HBr -------> C5H5NH(+) +Br(-)

HBr is limiting reactant and moles of C5H5N remaining= 0.00525-0.002625 =0.002625 moles

moles of C5H5NH+ formed= 0.002625, Volume after mixing= 25+12.5= 37.5ml

pOH = pKb + log [salt]/[base] , when salt/ base concentration ratio is taken, mole ratio same as concentration ratio

Concentration =

pOH= 8.75+log (0.002625/0.002625)= 8.75

pH= 14-8.75= 5.25

c)when 14ml of HBr is added. moles of HBr= 0.21*14/1000=0.00294, moles of salt formed= 0.00294 ( since HBr is limiting reactant), moles of pyridine remaining = 0.00525-0.00294=0.00231

pOH= 8.75+log (0.00294/0.00231)=8.85

pH= 14-8.85=5.15

d) when 25ml of Hbr is added, moles of HBr and pyridine are same and is equal to 0.00525 mole. This gives rise to 0.00525 moles of salt and hydrolysis of salt contributes to pH

Volume after mixing =25+25=50ml =50/1000 =0.05L

Concentration of salt, C= 0.00525/0.05=0.105 M

pH= 1/2{Pkw-pKb- log C} = 1/2{14-8.75-log(0.105)= 3.11

e) when 33ml of HBr is added, moles of HBr is excess by =0.21*33/1000-0.00525=0.00168

Concentration of HBr= 0.00168*1000/(33+25)=0.0289

HBr----> H+ +Br-

[H+] =0.0289

pH= -log(0.0289)=1.54

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