How to solve titration problems in analytical chemistry

Question

Can anyone please help me answer this?thank you.

In a titration of 50.00mL of 0.1000 M ethylamine with 0.10000 M HClO4, the titration error must be no more than 0.05mL. (What indicator can be chosen to realize this goal?) Just supply the pH value...

thank you

Answer 1

I'm Just Entering University At Chemistry Or Should I Say Chemical Engineering So I'll Try Help Out But Dont Fully Go With Me I Have A Different Method From Most Other People Ill Show You How I Done Titration Equations At School.

Okay This Is A Question From A Past Paper (Higher Chemistry 2004 PastPaper)

In a titration, a student found that an average of 16.7cm3 of Iron(II) Sulphate Solutation was needed to react completely with 25.0cm3 of 0.20mol-1 potassium permanganate solution.

The equation for the reaction is:

5Fe2+(aq) +MnO4-(aq) + 8H+(aq) ---> 5Fe3+(aq) +Mn2+(aq) + 4H20(l)

Now The Awnser To This Is

5Fe2+(aq) +MnO4-(aq) + 8H+(aq) ---> 5Fe3+(aq) +Mn2+(aq) + 4H20(l)

5mol           1mol  

Then You Can Say That:

VpMp (Permanganate)

--------- = 1 / 5

VsMs (Iron(II) Sulphate)

You Now Cross-Multiply These And Get:

VpMp (Permanganate)

--------- = 5 / 1

VsMs (Iron(II) Sulphate)                

You Can Arrange The Titration Equation To Give Ms (The Thing Your Looking For 5VpMp

                 ------------ = 5(25/1000x0.20) / 2(16.7/1000))

                     2Vs

Thats: 0.025 / 0.0167 which gives 1.5mol-1 which is the correct awnser on my awnser sheet (the working i got from a book just to make sure i didnt make any mistakes)

this is just an example to try help you solve your actual problem and its the only method i've ever been taught how to tackle problems like these i hope it at least helps give maybe a possible understand on how to solve your question =]

Answer 2

Since each is 0.1 M  then the EQ point should be at 100 mL   assuming a maximum of 0.05 mL past EQ ( your determination with indicator change is called endpoint )..Th total volume will be 100.05 mL   and the excess acid will be 0.05 mL x 0.1 N = 0.005 meq

the total concentration of the H+ at this 0.05 mL over true EQ   will be

0.005/100.05 = 4.998 X 10^-5 M  so pH at 0.05 mL over EQ will be

5 -log 4.998 = 4.3. The indicator methyl red is fully red at 4.4 which means the actual titration error will be slightly less than 0.05 mL