Calculate the pH during the titration of 10.00 mL of 0.400 M hypochlorous acid w

Question

Calculate the pH during the titration of 10.00 mL of 0.400 M hypochlorous acid with 0.500 M NaOH. First what is the initial pH (before any NaOH is added)? The Ka for HOCl is 3.0 x 10-8 M

PART B:How many mL of NaOH are added to reach the equivalence point?

PART C: What is the pH after 3.00 mL of NaOH are added?

PART D: What is the pH after 7.00 mL of NaOH are added?

PART E: What is the pH at the equivalence point?

PART F: What is the pH after 9.90 mL of NaOH are added?

PART G: What is the pH when half the acid has been neutralized?

PART H: What is the pH after 16.10 mL of NaOH are added?

Explanation / Answer

Ka of HOCl = 3*10^-8

so pKa=7.52

A)let the dissociation be x.so,

3*10^-8 = x^2/(0.4-x)

or x=1.095*10^-4

so pH=-log(x)

=3.96

B) let it be x mL.so,

x*0.5=10*0.4

=8 mL

C)pH=pKa+log(salt/acid)

= 7.52 + log(3*0.5/(10*0.4-3*0.5))

=7.29

D)pH=pKa+log(salt/acid)

=7.52 + log(7*0.0.5/(10*0.4-7*0.5))

=8.365

E)at the equivalence point,

pH=7+0.5pKa+0.5log(C)

=7+0.5*7.52+0.5*log(10*0.4/(10+8))

=10.4334

F)[OH-]=(9.9*0.5-10*0.4)/(10+9.9)

=0.0447

so pOH=1.35

so pH=12.65

G)at half the neutralisation point, pH=pKa

so pH=7.52

H)[OH-]=(16.1*0.5-10*0.4)/(10+16.1)

=0.155

so , pOH=0.809

so pH=13.19

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects