Question

2. When 30.00 mL of 0.1011 M HCl in 50 ml of Di water is titrated against 0.09889 M Naoui pH increases a. What is the volume (in mL) of NaOH required to reach the equivalence point and a pH of 7.007 b. Find the pH when the volume of NaOH added is 0.02 ml less than the volume required to reach the equivalence point. C. Find the pH when the volume of NaOH added is 0.01 mL less than the volume required to reach the equivalence point. d. Find the pH when the volume of NaOH added is 0.02 mL more than the volume required to reach the equivalence point. Note: some simple & valid approximation can facilitate the completion of c&d!

Answer 1

HCl and NaOH are strong acid and strong base respectively.

a) HCl + NaOH ------------> NaCl + H2O

At equivalence

mmoles of acid = mmoles of base

30x0.1011 = V x0.09889

Volume of base = 30.67 mL at equivalence

At equivalence as both are strong the pH = 7.00

b)when NaOh volume = 30.67-0.02 = 30.65 mL

HCl + NaOH ------------> NaCl + H2O

30x0.1011=3.033 30.65x0.09889=3.0309 0 0 initial mmoles

0.0021 0 3.0309 --- equilibrium

So the solution is still acidic and [H+] = mmoles/ volume = 0.0021/(30+30.65) =3.46x10-5

pH = -log 3.46x10-5

=4.4606

c) Volume of NaOH = 30.67-0.01 = 30.66 mL

HCl + NaOH ------------> NaCl + H2O

30x0.1011=3.033 30.66x0.09889=3.032 0 0 initial mmoles

0.001 0 3.032 --- equilibrium

So the solution is still acidic and [H+] = mmoles/ volume = 0.001/(30+30.66) =1.65x10-5

pH = -log 1.65x10-5

= 4.7829

d)

when NaOh volume = 30.67-0.02 = 30.65 mL

HCl + NaOH ------------> NaCl + H2O

30x0.1011=3.033 30.69x0.09889=3.0349 0 0 initial mmoles

0 0.0021 3.0309 --- equilibrium

So the solution is basic  and [OH-] = mmoles/ volume = 0.0021/(30+30.65) =3.46x10-5

pOH = -log 3.46x10-5

=4.4606

and pH = 14-pOH = 14-4.4606 =9.5394