a) initial pH
pkb = - logkb
= -log(7.4*10^-6)
= 5.13
pOH = 1/2(pkb-logC)
= 1/2(5.13-log0.88)
= 2.593
pH = 14-poH
= 14-2.593
= 11.407
b)after addition of HCl
no of mol of weakbase taken = 49.1*0.88 = 43.21 mmol
no of mol of HCl = 6*1.8 = 10.8 mmol
pH = 14 - (pkb + log(salt/base))
= 14 - (5.13+log(10.8/(43.21-10.8)))
= 9.35
c) after addition of HCl
no of mol of weakbase taken = 49.1*0.88 = 43.21 mmol
no of mol of HCl = 12*1.8 = 21.6 mmol
pH = 14 - (pkb + log(salt/base))
= 14 - (5.13+log(21.6/(43.21-21.6)))
= 8.87
d) after addition of HCl
no of mol of weakbase taken = 49.1*0.88 = 43.21 mmol
no of mol of HCl = 18*1.8 = 32.4 mmol
pH = 14 - (pkb + log(salt/base))
= 14 - (5.13+log(32.4/(43.21-32.4)))
= 8.39
e) after addition of HCl
no of mol of weakbase taken = 49.1*0.88 = 43.21 mmol
no of mol of HCl = 24*1.8 = 43.2 mmol
reaches to equivalencepoint
concentration of salt = 43.2/(49.1+24) = 0.591 M
pH = 7-1/2(pkb+logC)
= 7-1/2(5.13+log0.591)
= 4.55
f) after addition of HCl
no of mol of weakbase taken = 49.1*0.88 = 43.21 mmol
no of mol of HCl = 30*1.8 = 54 mmol
concentration of excess HCl = (54-43.21)/(54+49.1) = 0.105 M
pH = -log(H3O+)
= -log(0.105)
= 0.98
Show how to get the following pH’s (on left) from the titration problem A 1.8 M...
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Which of the following titration curves corresponds to the
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Question Completion Status: QUESTION 4 Which of the following titration curves corresponds to the titration of a weak base with HCI? Hd A0006 5 10 15 20 25 30 35 Volume of HCl added (mL) 14+ B Ob. Ś 10 is zo 25 30 35 Volume of HCl added (ml) 14+ pH No 0 5 5 s 10 is zo 25 30 35 Volume...
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Consider the titration of 60.0 mL of 0.0400 M C H5NH2 (a weak base; Kb = 4.30e-10) with 0.100 M HBr. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL (C) 12.0 mL (b) 6.0 mL pH = pH = x pH...
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