Question

A beaker with 1.50

Answer 1

Let the molarity of acetic acid be a

molarity of conjugate base be b

given

total molarity = 0.1

so

a + b = 0.1

also

for a acidic buffer

pH = pKa + log [ conjugate base / acid ]

so

5 = 4.76 + log [ b / a ]

b/a = 1.7378

but

b = 0.1 - a

so

( 0.1 - a ) / a = 1.7378

0.1 /a = 2.7378

solving we get

a = 0.0365

b = 0.1 - 0.0365

b = 0.0635 M

now given 150 ml of buffer

we know that

moles = molarity x volume (L)

so

moles of acetic acid = 0.0365 x 150 x 10-3 = 5.475 x 10-3

moles of conjugate base = 0.0635 x 150 x 10-3 = 9.525 x 10-3

moles of HCl added = 0.36 x 5.7 x 10-3 = 2.052 x 10-3

now the reaction is given by

HCl + conjuagte base (CH3COO- ) ----> acetic acid ( CH3COOH )+ Cl-

from the above reaction

moles of conjugate base reacted = moles of HCl added = 2.052 x 10-3

moles of conjugate base remaining = 9.525 x 10-3 - 2.052 x 10-3 = 7.473 x 10-3

moles of acetic acid formed = moles of conjugate base reacted = 2.052 x 10-3

moles of acetic acid present = 5.475 x 10-3 + 2.052 x 10-3 = 7.527 x 10-3

now

new pH = pKa + log [ conjugate base / acid ]

we also know that

molarity = moles / volume (L)

as the final volume is same for both

ratio of molarity = ratio of moles

so

new pH = 4.76 + log [ 7.473 x 10-3 / 7.527 x 10-3 ]

new pH = 4.75687

so

change in pH = 5 - 4.75687

change in pH = 0.243

so the pH is decreased by 0.24