Question

potentiometric titration help!

Answer 1

I need to find the % of my unknow.

Answer 2

@ganeshie8

Answer 3

i dont want to write up my whole lab so I kinda need to know what I should include

Answer 4

So far i have \[V_{eq}=34.71 mL\] the titration is between a standard NaOH with an ionic strength adjustment with NaCl and a unknown containing KHP. I need to find %KHP in the original sample.

Answer 5

First you need to find out the molecular weight of your unknown. Did you find that already?

Answer 6

I'm assuming they're looking for w/w%?

Answer 7

yeah that is what I'm looking for.

Answer 8

well I have the weight of my unknown KHP \[.5005g\]
It was dissolved in a 250 mL volumetric flask to the mark.

Answer 9

Also 10.53 NaCl mL was added to the unknown solution before dilution.
10.00 mL was added to my primary standard NaOH

Answer 10

Molarity of the NaOH solution was .0075 M

Answer 11

Ok, let me see if I can remember, I haven't taken quant in a few years. But I think to find w/w%, you need to know the mass of sample and its weight.

Answer 12

You also need to know the equivalence point, I think that will tell you the equivalence weight.

It should be something like \(\sf KHP~mass \times \frac{1~NaOH}{1~KHP}\times \frac{1~L}{molarity}\times \frac{1}{mL~to~reach~equivalence}\)

Answer 13

I'm assuming the reaction is a 1:1 ratio of KHP with the base.

Answer 14

34.71 mL is when the titration reaches equivalence I've been working backwards.
\[34.71 mL NaOH*\frac{1L}{1000 mL NaOH}*\frac{.0075 moles NaOH}{L}*\frac{1moleKHP}{1moleNaOH}\]
\[\frac{204.2 gKHP}{1mole KHP}=.0532656g KHP\]
Then i did this:
\[\frac{.0532656 gKHP}{.5005gunknown}*100=11% KHP\]
I got it wrong so i don't know what to do.

Answer 15

*11% KHP

Answer 16

@abb0t

Answer 17

@aaronq

Answer 18

@Preetha