Question

Your acid has a pka of 6.45 and a concentration of 0.079M. You are given 40.8mL of this acid, what is the pH of this initial solution? Calculate the pH of the solution after adding 6.8, 15.3, 27.9, and 33.7 mL of 0.100M NaOH. At the equivalence point, what volume of base is added and what is the pH?

Answer 1

we know that

pKa = -log Ka

6.45= -log Ka

Ka = 3.548 x 10-7

for weak acids

[H+] = sqrt ( Ka x C)

so

[H+] = sqrt ( 3.548 x 10-7 x 0.079)

[H+] = 1.674 x 10-4

now

pH = -log [H+]

so

pH = -log 1.674 x 10-4

so

pH = 3.776

so

initially the pH is 3.776

2)

now

6.8 ml of NaOH is added

we know that

moles = conc x volume (L)

so

moles of NaOH added = 0.1 x 6.8 x 10-3 = 0.68 x 10-3

moles of acid = 0.079 x 40.8 x 10-3 = 3.2232 x 10-3

now

the reaction is

HA + NaOH ---> NaA + H20

we can see that

moles of HA reacted = moles of NaOH added = 0.68 x 10-3

so

moles of HA remaining = 3.2232 x 10-3 - 0.68 x 10-3 = 2.5432 x 10-3

moles of NaA formed = moles of NaOH added = 0.68 x 10-3

now

pH = pKa + log [ NaA / HA]

so

pH = 6.45 + log [0.68 x 10-3 / 2.5432 x 10-3 ]

pH = 5.877

so

the pH after addition of 6.8 ml NaOH is 5.877

3)

now

15.3 ml of NaOH is added

we know that

moles = conc x volume (L)

so

moles of NaOH added = 0.1 x 15.3 x 10-3 = 1.53 x 10-3

moles of acid = 0.079 x 40.8 x 10-3 = 3.2232 x 10-3

now

the reaction is

HA + NaOH ---> NaA + H20

we can see that

moles of HA reacted = moles of NaOH added = 1.53 x 10-3

so

moles of HA remaining = 3.2232 x 10-3 - 1.53 x 10-3 = 1.6932 x 10-3

moles of NaA formed = moles of NaOH added = 1.53 x 10-3

now

pH = pKa + log [ NaA / HA]

so

pH = 6.45 + log [1.53 x 10-3 / 1.6932 x 10-3 ]

pH = 6.406

so

the pH after addition of 15.3 ml NaOH is 6.406

4)

now

27.9 ml of NaOH is added

we know that

moles = conc x volume (L)

so

moles of NaOH added = 0.1 x 27.9 x 10-3 = 2.79 x 10-3

moles of acid = 0.079 x 40.8 x 10-3 = 3.2232 x 10-3

now

the reaction is

HA + NaOH ---> NaA + H20

we can see that

moles of HA reacted = moles of NaOH added = 2.79 x 10-3

so

moles of HA remaining = 3.2232 x 10-3 - 2.79 x 10-3 = 0.4332 x 10-3

moles of NaA formed = moles of NaOH added = 2.79 x 10-3

now

pH = pKa + log [ NaA / HA]

so

pH = 6.45 + log [2.79 x 10-3 / 0.4332 x 10-3 ]

pH = 7.26

so

the pH after addition of 27.9 ml NaOH is 7.26

5)

now

33.7 ml of NaOH is added

we know that

moles = conc x volume (L)

so

moles of NaOH added = 0.1 x 33.7 x 10-3 = 3.37 x 10-3

moles of acid = 0.079 x 40.8 x 10-3 = 3.2232 x 10-3

now

the reaction is

HA + NaOH ---> NaA + H20

we can see that

moles of NaOH reacted = moles of HA = 3.2232 x 10-3

so

moles of NaOH remaining = 3.37 x 10-3 - 3.2232 x 10-3 = 0.1468 x 10-3

total volume = 33.7 + 40.8 = 74.5 ml

now

[NaOH] = 0.1468 x 10-3 x 1000 / 74.5 = 1.97 x 10-3

now

NaOH ---> Na+ + OH-

so

[OH-]= [NaOH] = 1.97 x 10-3

pOH = -log [OH-]

so

pOH = -log 1.97 x 10-3 = 2.705

so

pH = 14 - pOH

pH = 14 - 2.705

pH = 11.295

so

after addition of 33.7 ml NaOH , pH is 11.295

6)

at equivalence point

Ma x Va = Mb x Vb

so

0.079 x 40.8 = 0.1 x Vb

Vb = 32.232 ml

so

32.232 ml of base should be added to get to equivalence point

now

moles of NaA formed = 3.2232 x 10-3

total volume = 32.232 + 40.8 = 73.032 ml

so

[NaA] = 3.2232 x 10-3 x 1000 / 73.032

[NaA] = 0.044

now

NaA is a weak base

for

weak bases

[OH-] = sqrt ( Kb x C)

[OH-]= sqrt ( Kw x C / Ka)

so

[OH-] = sqrt ( 10-14 x 0.044 / 3.548 x 10-7)

[OH-] = 3.527 x 10-5

pOH = -log 3.527 x 10-5

pOH = 4.45

pH = 14 - 4.45

pH = 9.55

so

pH at equivalence point is 9.55