Question

Can someone check my work and tell me what I did wrong? Thankyou!

Answer 1

I performed a potentiometric titration:
\[V_{eq}=34.71 mL\] volume at equivalency point
\[.5005g\]
is the weight of my unknown
Also 10.53 NaCl mL was added to the unknown solution before dilution. 10.00 mL was added to my primary standard NaOH

Molarity of the NaOH solution was .0075 M

Answer 2

Here is what I did and it was marked wrong... help me figure out why:

Answer 3

\[34.71 mL NaOH*\frac{1L}{1000 mL NaOH}*\frac{.0075 moles NaOH}{L}*\frac{1moleKHP}{1moleNaOH}\]
\[\frac{204.2 gKHP}{1mole KHP}=.0532656g KHP\]
\[\frac{.0532656 gKHP}{.5005gunknown}*100=11percent KHP\]

Answer 4

@Australopithecus do you think you can help me?

Answer 5

What were all the steps you performed? We need to know exactly what you did otherwise no one will be able to help you. Sounds like you used an internal standard as well but I cant be sure unless I know what you did exactly.

Answer 6

So you want to know the steps of my equation? or my procedure?

Answer 7

ok well I prepared a primary standard solution of .0075 M NaOH
10.00 mL 3.56 M NaCL and 25 mL .0075 M NaOH was added to a 250 mL volumetric flask and diluted to mark.
An unknown (.5000g) was added to a 250 mL flask plus about 10.53 mL of 3.56 M NaCl and diluted to mark.
I use 50 mL of the unknown solution to a 250 mL beaker and inserted a pH meter. I titrated the NaOH solution into the beaker and recorded multiple pH and volumes.
So I got what volume the NaOH is need to get to the equivalency point.
using this info I did that equation and it didn't work...

Answer 8

@Australopithecus is this what you needed to help me?

Answer 9

@Kainui think you could help me out?

Answer 10

NaOH diluted concentration
(25mL/250mL)*0.0075M

NaCl diluted concentration
(10mL/250mL)*3.56M

Did the lab come with a reaction equation or the compound in the unknown you were titrating against?

Answer 11

I guess the unknown is KHP

Answer 12

it didn't come with any equations =/

Answer 13

the NaCl was used to adjust the ionic strength... do I need to include it into the equation?

Answer 14

Ok,

so,

KHC8H4O4(aq) + NaOH(aq) -> KNaC8H4O4(aq) + H2O(l)

so it appears to be 1 to 1 NaOH to KHP

34.71mL is the equivalence point

Then it is as simple as finding moles of NaOH in 34.71mL,
(25mL/250mL)*0.0075M

use them in the formula:

Molarity = Moles/liters

Then you have moles of NaOH which = moles of KHP

Then get the molecular mass of KHP use the formula

mole = grams/molecular mass

solve for grams

% KHP in unknown = (Grams KHP/Grams unknown)*100

Answer 15

But I'm supposed to use the volume it takes to reach the equivalency point.

Answer 16

34.71mL is the volume it takes to reach equivalence

Answer 17

You take the volume of NaOH find the moles in that volume then realize that the reaction is 1 to 1 so the moles in 34.71mL of NaOH = the moles of KHP in the sample

Answer 18

so..
\[(\frac{25mL}{250mL})*.0075M=.0006MNaOH\]
\[34.71 mL NaOH*\frac{1L}{1000 mL NaOH}*\frac{.0006 moles NaOH}{L}*\frac{1moleKHP}{1moleNaOH}\]
\[*\frac{204.2 gKHP}{1mole KHP}=.0042527g KHP\]
\[\frac{.0042527 gKHP}{.5005gunknown}*100=.85percent KHP\]
but my unknown was diluted too how would I do that?

Answer 19

oh sorry I made a mistake new instructions:

KHC8H4O4(aq) + NaOH(aq) -> KNaC8H4O4(aq) + H2O(l)

so it appears to be 1 to 1 NaOH to KHP

34.71mL is the equivalence point

Then it is as simple as finding moles of NaOH in 34.71mL,
(25mL/250mL)*0.0075M

use them in the formula:

Molarity = Moles/liters

**NEW steps start here ***

Then you have moles of NaOH which = moles of KHP in 50mL

Calculate Molarity of KHP in unknown solution (concentration remains constant)

Volume = 0.050L
moles of KHP already determined

Molarity = Moles/Liters

Once you have molarity use the formula again with a volume of 0.250L and solve for moles, this will give you the total moles present in the unknown solution

Alternatively you could just multiply moles of NaOH by (250mL/50mL)

**New steps end here***

Then get the molecular mass of KHP use the formula:

mole = grams/molecular mass

solve for grams

% KHP in unknown = (Grams KHP/Grams unknown)*100

Answer 20

still confused.

Answer 21

I think I go it now.. I guess we will see thank you for your help.

Answer 22

I don't how I could be wrong

1. Figure out moles of NaOH make sure to account for dilutions
2. Realize 1 to 1 reaction so moles NaOH = Moles KHP
3. Concentration is constant in regards to taking an aliquot of a larger volume so you can just figure out the concentration in the smaller volume 50mL using moles of KHP then use the formula again with the full volume 250mL or just by multiplying moles by 250/50 because 50mL contains 1/5th of the moles in the original volume 250mL.
4. Use moles = g/molecular mass of KHP to find grams of KHP in unknown
5. Apply (mass of KHP/Mass of sample)*100 = % KHP in unknown.

Problem Solved