volumetric analysis question (1 Viewer)

[Farmerism]

im havin trouble writing the equation for this problem

find the molarity of potassium hydroxide if 47.8mL of it reacts completly with 0.26g of oxalic acid (COOH)2.2H2O

i dont get the "(COOH)2.2H2O" part.. what exactly do you do with it

 

[thejosiekiller]

hint:
find the molecular weight of oxalic acid

then divide the weight of the sample by its MW giving you the number of moles of acid reacting

the rest should be ok from there

 

[Nodice]

Write oxalic acid as: H2C2O4 this way its much easier to see that it is diprotic, and you get a general idea of how it reacts.

H2C2O4 + KOH ---> ....

 

[Farmerism]

Write oxalic acid as: H2C2O4 this way its much easier to see that it is diprotic, and you get a general idea of how it reacts.

H2C2O4 + KOH ---> ....

um so how exactly does it react?

i asked my mum this and she said "according to the general word equation:

Acid + Base -> salt + water

the H+ from (COOH)2 [from (COOH)2.2H2O] reacts with the OH- from KOH to form water, the remainder is the salt, so the equation would have to be...

2KOH + (COOH)2.2H2O -> (COOK)2 + 4H2O"

so... what does everyone else think? (mum hasnt done chem since uni and so shes a little rusty on these things, but i appreciate her help! i just need a second opinion)

 

[Farmerism]

okay well i just used mum's equation to determine the mole ratio and molarity etc and so im guess that her logic was right

my working out, just for the record was:

2KOH + (COOH)2.2H2O -> (COOK)2 + 4H2O

data:
KOH - 2moles , Vol.=47.8mL , M=?
(COOH)2.2H2O - 1 mole, mass=0.26g, FW=126 amu

the math:
n(COOH)2.2H2O) = mass/ FW = 0.26/ 126 = (13/6300)

since 2 moles of KOH react with 1 mole of (COOH)2.2H2O

nKOH = 2 x (13/6300) = 13/3150

hence, molarity = n/v = (13/3150) / 0.0478 = 0.09molL^-1

this is right, according to excel..........................

an explanation helps guys, not just, "do this because thats how youre suppose to do it"

and

hint:
find the molecular weight of oxalic acid

then divide the weight of the sample by its MW giving you the number of moles of acid reacting

the rest should be ok from there

how can you do that when you dont know the equation; the mole ratio that reactants react with?

and

Write oxalic acid as: H2C2O4 this way its much easier to see that it is diprotic, and you get a general idea of how it reacts.

H2C2O4 + KOH ---> ....

it doesnt matter whether its diprotic, its a volumetric question, not a pH question, you can work out concentrations using the c=n/v formula..
and i dont think you could get the right products if you started the reaction that way.

hope this enlightens people

 

[thejosiekiller]

okay well i just used mum's equation to determine the mole ratio and molarity etc and so im guess that her logic was right

my working out, just for the record was:

2KOH + (COOH)2.2H2O -> (COOK)2 + 4H2O

data:
KOH - 2moles , Vol.=47.8mL , M=?
(COOH)2.2H2O - 1 mole, mass=0.26g, FW=126 amu

the math:
n(COOH)2.2H2O) = mass/ FW = 0.26/ 126 = (13/6300)

since 2 moles of KOH react with 1 mole of (COOH)2.2H2O

nKOH = 2 x (13/6300) = 13/3150

hence, molarity = n/v = (13/3150) / 0.0478 = 0.09molL^-1

this is right, according to excel..........................

an explanation helps guys, not just, "do this because thats how youre suppose to do it"

and



how can you do that when you dont know the equation; the mole ratio that reactants react with?

e

well you said in your first post it reacts completely, so theoritically n moles of the acid will react completely...i just skipped finding the mole ratio cause thats pretty easy when looking at the equation itself

 

[Farmerism]

um, the problem i had initially was writing the chemical equation, so that the products were correct.. not the calculating part. so yeah when you say

well you said in your first post it reacts completely, so theoritically n moles of the acid will react completely...i just skipped finding the mole ratio cause thats pretty easy when looking at the equation itself

it didnt really help with anything.....hehe

but its all cool now!

i have my half yearlys for chem and music tomorrow!!...........:'( im gonna fail

 

[thejosiekiller]

ah sry- the equation is the easy part and in most cases given to you already

but i didnt do hsc chem

good luck