I\'ve been struggling so much with these damn questions because I dont have the

Question

I've been struggling so much with these damn questions because I dont have the Ka or pKa to work with...so im lost as to what i have to do. teacher does not provide us with chart on the exam for pka or ka values either.

Question:

Calculate the pH at each point indicated in the titration of 100.00 mL of 0.120 M HClO4 with 0.0375 M NaOH.

(Hint: Watch out for having to do one via systematic treatment of equilibrium)

a) Addition of 85.0 mL NaOH solution

b) Addition of 100.00 mL NaOH solution

c) Addition of 160.00 mL NaOH solution

d) Addition of 320.00 mL NaOH solution

e) Addition of 323.00 mL NaOH solution

Explanation / Answer

HClO4 is a strong acid.so,

a)HClO4 + NaOH ---> NaClO4 + H2O

so [H+]=(100*0.12-0.0375*85)/(100+85)

=0.0476

so pH=-log(H+)

=1.32

b) [H+]=(100*0.12-0.0375*100)/(100+100)

=0.04125

so pH=-log(H+)

=1.38

c)[H+]=(100*0.12-0.0375*160)/(100+160)

=0.02308

so pH=1.637

d)[H+]=(100*0.12-0.0375*320)/(100+320)

=0.0

so equivalent point.

so pH=7 (only due to water

e)[OH-]=(0.0375*323-100*0.12)/(100+323)

=2.66*10^-4

so pOH=3.575

so pH=14-pOH

=10.42

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