Question

In a titration of 20 mL of 0.20 M HX with a NaOH solution the equivalence point volume of base is 30 mL. If the PH is 3.74 @ 15 mL of base what is the pH @ 10 mL of added base?

Answer 1

millimoles of HX = 0.20 x 20 /1000 = 4 x 10^-3

base volume at equivalence point = 30 mL

at equivalence point

moles of acid = moles of base

4 x 10^-3 = M x 0.03

Molarity of NaOH = 0.133 M

the PH is 3.74 @ 15 mL of base

moles of base = 0.133 x 15 /1000 = 2 x 10^-3

HX       + NaoH   ---------------> NaX +   H2O

4 x 10^-3    2x10^-3                         0           0

2 x 10^-3        0                            2x10^-3     2x10^-3

pH = pKa + log[salt/acid]

3.74 = pKa + log[2x10^-3/2x10^-3]

pKa = 3.74

pH @ 10 mL of added base

moles of base = 0.133 x 10 / 1000 = 1.33 x 10^-3

HX       + NaoH   ---------------> NaX +   H2O

4 x 10^-3    1.33 x10^-3                         0           0

2.67 x 10^-3        0                            1.33 x10^-3   1.33x10^-3

pH = pKa + log[salt/acid]

pH = 3.74 + log[1.33 x10^-3 / 2.67x10^-3]

pH = 3.44