Back to basics (1 Viewer)

[babydoll_]

Ok I need some help with chem... I dont know what I'm doing wrong

Titration: Potassium hydrogenphthalate against sodium hydroxide
Potassium hydrogenphthalate: KHC8H4O4 . Molar mass: 204.22
Sodium hydroxide: NaOH . Molar mass: 39.998

Potassium hydrogenphthalate (ill just call it KHP) was made into a standard solution of approximately 0.1M.

Mass of KHP used: 5.05

Number of moles of KHP used: 5.05/204.22 = 0.0247 mol

25.0 mL of this was used to titrate NaOH of an unknown concentration.

Average titre = 24.4.mL

Now, # of moles of KHP = moles of NaOH

So concentration of NaOH = n/v
which is 0.0247/0.0244 = 1.012M

is this correct??? cos i think im wrong but im not sure where

 

[Xayma]

Based on the fact you said 0.1M I assume you used 250mL which means you miscalculated the number of moles used. You only used 1/10 of that number of moles (since you only used 25mL). so your answer is 10x too big

 

[babydoll_]

OH ok i sorta get it

Ta muchly ^^

now wait while i hit you with another question

 

[babydoll_]

This might be out of your scope but I'm having some serious trouble with this

1/ A solution was prepared by dissolving 4.119g of potassium phosphate (K3PO4) (mm=212.27g mol^-1) in sufficient water to prepare 323.3mL of solution. Calculate the concentration of potassium ions and the concentration of phosphate ions.

2/ Calculate the volume of this solution which would contain 0.00212 mol of phosphate ions.

i dont know where to start..

 

[Xayma]

Im assuming it completly dissociates into its ions.

n_K3PO4=4.119/212.27
=0.0194mol

By stoich
n_K=3n_K3PO4=0.05821mol

Similary n_PO4=0.0194mol

now c=n/v
c_K=0.05821/.3233=0.1800mol L^-1
c_PO4=0.0194/.3233=0.0600mol L^-1

2. c=n/v
therefore v=n/c
v=0.00212/0.06002
=0.03532L
=35.32mL

 

[babydoll_]

ooh thank you very much