[Twickel]

Hi
I am having trouble with determining the dilution factors.

I have an unknown of manganese given to me. I have to adjust this solution to 100mL, I then pipette three 20mL amounts of that 100mL into three 250mL conical flasks.

To each solution I dilute up to roughly 100Ml, add 10mL of phosphoric acid and 0.3g of potassium periodate. It is then heated, this heatd solution is colled and then transferred with rinsing to another 250mL volumetric flask. This solution is adjusted to 250mL


Know, after I get the abs and determines concentration is this correct?

Number of moles = concentration x 250mL/ 20mL.
Then multiply by 5 to account for the pippeting of 20mL from the 100mL

[Borek]

Looks OK to me.

[Twickel]

Thank goodness.

And the standard solutions, I pipette 10,20 and 25mL. I dilute each volume to roguhly 100mL add the acid and periodate, but this 100mLis not a dilution its just to help with the reaction.
It is then diluted to 250mL so concentration for the x axis will be

0.00320Mx10/250=x 0.00320x20/250=x2 and so on.

Btw, how does MnCl2.4H2O dissociate?
Is it a1:1?

[Twickel]

Looks OK to me.

Actually is the number of moles equal to (cx250) x5 not cx250/20

[Twickel]

From my graph I got concentration in unknown to be 2.1669x10^-4 M.


Now how do I figure out mass of MnCl2.4H20?

[Borek]

You convert each Mn²⁺ to MnO₄^-, 1:1.

[Twickel]

I am not sure how to figure out the number of moles though.

What I did was.
n= c x v ( 2.6619x10^-4 x 0.25) x 5 ( because we  only pipetted 20 of 100mL).

then what ever n was x molar mass of Mncl2.4H20  its wrong though.... 

[Borek]

Hard to say anything without seeing all numbers. What you wrote so far looks at least logical in the context given.