Question

Concentraion of NaOH = 0.1023 M

Volume of acetic acid = 20 ml

Volume of NaOH required to reach equivalent point = 20.80ml

Calculate the anticipated PH after 9.0ml of the standard NaOH have been added to the acetic acid solution. Show all calculations and explain/justify any assumptions.

Answer 1

we know that

at equivalence point

Ma x Va = Mb x Vb

so

Ma x 20 = 0.1023 x 20.8

Ma = 0.106392

so

concentration of CH3COOH = 0.106392

now

moles = molarity x volume (L)

so

moles of CH3COOH present = 0.106392 x 20 x 10-3 = 2.12784 x 10-3

now

moles of NaOH added = 0.1023 x 9 x 10-3 = 0.9207 x 10-3

now

the reaction is

CH3COOH + NaoH ---> CH3COONa + H20

we can see that

moles of CH3COOH reacted = moles of NaOH added = 0.9207 x 10-3

now

new moles of CH3COOH = 2.12784 x 10-3 - 0.9207 x 10-3 = 1.20714 x 10-3

now

moles of CH3COONa formed = moles of NaOH added = 0.9207 x 10-3

now

CH3COONa and CH3COOH form a buffer solution

for buffers

pH = pKa + log [salt / acid

so

pH = pKa + log [CH3COONa / CH3COOH]

pH = 4.76 + log [0.9207 x 10-3 / 1.20714 x 10-3 ]

pH = 4.64

so

the pH is 4.64