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Find the pH during the titration of 20.00 mL of 0.2830 M benzoic acid, C6H5COOH (Ka =...

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Chemistry

Find the pH during the titration of 20.00 mL of 0.2830M benzoic acid, C6H5COOH(Ka = 6.3 ✕ 10-5), with 0.2830M NaOH solution after the following additions oftitrant.

(a)    0 mL


(b)    10.00 mL


(c)    15.00 mL


(d)    19.00 mL


(e)    19.95 mL


(f)    20.00 mL


(g)    20.05 mL


(h)    25.00 mL

Answer & Explanation Solved by verified expert
4.1 Ratings (598 Votes)

pKa of the acid = 4.2

Amount of the base required for neutralization = 20 mL

so,

a)let the amount of dissociation be x.so,

6.3*10^-5 = x^2/(0.283-x)

or x=0.00419

so pH=2.377

b)a buffer solution will be formed.

so, ph=pKa+log(salt/acid)

=4.2 + log(10*0.283/(20*0.283-10*0.283))

=4.2

c)

a buffer solution will be formed.

so, ph=pKa+log(salt/acid)

=4.2 + log(15*0.283/(20*0.283-15*0.283))

=4.677

d)

a buffer solution will be formed.

so, ph=pKa+log(salt/acid)

=4.2 + log(19*0.283/(20*0.283-19*0.283))

=5.478

e)

a buffer solution will be formed.

so, ph=pKa+log(salt/acid)

=4.2 + log(19,95*0.283/(20*0.283-19.95*0.283))

=6.8

f)concentration of the salt formed = 0.283/2

=0.1415

so,

since this is a salt of a weak acid and a strong base,

pH=7 + 0.5pKa+0.5*log(C)

=7+0.5*4.2 + 0.5*log(0.1415)

=8.675

g) there is excess of base present.

[OH-]=0.05*0.285/(20+20.05)

=3.55*10^-4

so pH=10.55

f)

there is excess of base present.

[OH-]=5*0.285/(20+25)

=0.031667

so pH=12.5
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