Find the pH during the titration of 20.00 mL of 0.1930 M butanoic acid, CH3CH2CH
ID: 956669 • Letter: F
Question
Find the pH during the titration of 20.00 mL of 0.1930 M butanoic acid, CH3CH2CH2cooH (Ka 1.54 x 10 5), with 0.1930 M H solution after the following additions of titrant. (a) 0 mL. 2.76 (b) 10.00 mL 4.8 (c) 15.00 mL 5.42 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. (d) 19.00 mL 5.46 Your response differs from the correct answer by more than 10%. Double check your calculations. (e) 19.95 mL (f) 20.00 mL. (g) 20.05 mL (h) 25.00 mLExplanation / Answer
a)
pka of CH3CH2COOH = -LOGKa = -log(1.54*10^(-5)) = 4.81
pH = 1/2(pka-logC)
= 1/2(4.81-log0.193) = 2.76
b)
No of mol of CH3CH2COOH = 20/1000*0.193 = 0.00386 mol
No of mol NaOH added = 10/1000*0.193 = 0.00193 mol
it is at half equivalence point
pH = pka = 4.81
c)
No of mol of CH3CH2COOH = 20/1000*0.193 = 0.00386 mol
No of mol NaOH added = 15/1000*0.193 = 0.0029 mol
pH = pka+log(salt/acid)
= 4.81 + log(0.0029/(0.00386-0.0029))
= 5.29
d) 19 ml
No of mol of CH3CH2COOH = 20/1000*0.193 = 0.00386 mol
No of mol NaOH added = 19/1000*0.193 = 0.00367 mol
pH = pka+log(salt/acid)
= 4.81+log(0.00367/(0.00386-0.00367))
= 6.095
e) 19.95 ml
No of mol of CH3CH2COOH = 20/1000*0.193 = 0.00386 mol
No of mol NaOH added = 19.95/1000*0.193 = 0.00385 mol
pH = pka+log(salt/acid)
= 4.81+log(0.00385/(0.00386-0.00385))
= 7.39
f) 20 ml
Concentration of salt = 20*0.193/40 = 0.0965 M
pH = 7+1/2(pka+logC)
= 7+1/2(4.81+log0.0965)
= 8.9
g)
excess NaOH concentration = (20.05-20)/40.05*0.193
= 0.00024
pOH = -log0.00024 = 3.62
pH = 14-3.62 = 10.38
h)
excess NaOH concentration = (25-20)/45*0.193
= 0.0214
pOH = -log0.0214 = 1.67
pH = 14-1.67 = 12.33
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