100 ml of 1 M acetic acid (pK a 4.76) are placed in a flask. The following volum

Question

100 ml of 1 M acetic acid (pKa 4.76) are placed in a flask.  The following volumes of 1 M KOH are added, and the total volume is adjusted to one liter with water.

a) 0 ml, pH = 2.88                    d) 25 ml                       g) 90 ml                       j) 101 ml

b) 2 ml (answer 3.07)                        e) 50 ml                       h) 98 ml

c) 10 ml                                   f) 75 ml                        i) 100 ml, pH = 8.88

I don't know where to start, for example how would you get a pH of 3.07 for b

Explanation / Answer

a) For 0 ml, pH = pKa/2 = 2.38

b) Buffer as above, Salt = 2 m-mole; Acid remaining = 98 m-moles

pH = pKa + log(salt/acid) = 3.05

c) Salt = 10 m-mole, Acid = 90 m-mole

pH = 3.786

d) For 25 ml, The solution will act as buffer,

Salt formed = 25 m-moles ; acid remaining = 75 m-mole

pH = pKa + log(salt/acid) = 4.26

e) Salt = 50 m-mole ; Acid = 50 m-mole

pH = pKa = 4.74

f) salt = 75 m-mole; Acid = 25 m-mole

pH = 5.22

g) For 90 ml, Same, Buffer.

Salt = 90 m-mole ; Acid = 10 m-mole

pH = pKa + log(salt/acid) = 5.69

h) salt = 98 m-mole ; acid = 2 m-mole

pH = 6.43

i) Salt = 100 m-mole; No acid, Hence, hydrolysis of salt will take place.

CH3COONa + H2O -----------> CH3COOH + OH- + Na+

[OH-] = sqrt(Kh*c) ; Kh = Hydrolysis constant = Kw/Ka = 10^-14/(1.8*10^-5) = 5.56*10^-10

c = 100m-mole/1 L = 0.1 M

[OH-] = 7.45*10^-6 M

pOH = 5.13 ; Hence, pH = 14-pOH = 8.87

j) For 101 ml, Base is more and since it is strong, it will dominate,

OH- = 1ml*1M = 1 m-mole (Rest 100 ml base will be neutrilized)

[OH-] = 1*10^-3 / 1L = 10^-3 M

pOH = 3 ; Hence, pH = 14-pOH = 11

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