Hard question, need assistance... (1 Viewer)

[WouldbeDoctor]

p 166

q26.

a.) Oxalic acid dihydrate, (cooh)2.2h20 can be used as a primary standard for standardising alkali solution. 0.291 g diprotric oxalic acid required 18.2 ml of a potassium hydroxide solution for exact neutralisation. calculate the molarity of the hydroxide solution.

 

[WouldbeDoctor]

Okay, this looks daunting but isn’t so hard if you break it down.

1. Convert the mass of oxalic acid to moles:<O</O
0.291 g (COOH)<SUB>2</SUB>.2H<SUB>2</SUB>O = 0.291/[2{12.01 + 2 (16.00) + 1.008} +2{2(1.008) + 16.00}]= 0.0023 mol (4 sf)
<O</O
2. Write the neutralization equation involved:<O</O
(COOH)<SUB>2 </SUB>+ 2KOH --> K<SUB>2</SUB>C<SUB>2</SUB>O<SUB>4</SUB> + 2H<SUB>2</SUB>O
<SUB></SUB><O</O
3. From the equation, figure out how many moles of KOH would react with the amount in step 1:<O</O
Since 1 mol oxalic acid reacts with 2 mol KOH, 0.0032 mol oxalic acid would react with 0.0023 x 2 = 0.0046 mol KOH (retaining calculator’s decimal memory from step 1- you shouldn’t round off until the last step.)
<O</O
4. The amount of KOH from step 3 is dissolved in the 18.2 mL mentioned in the question. From this we can calculate the molarity of the solution:
m<O</Oolarity = no. of moles/volume = 0.0046/0.0182 = 0.254 mol L<SUP>-1</SUP> KOH (3 sf)<O</O
<O</O

Voila! Hope that helped =)<O</O

Thanks, your answer is correct, however, the reason why I couldn't do it, is because I didn't know the eqn, products... Your chemistry is excellant, could you explain how you got k2c2o4 because this is the trickiest bit. How do I find the products of these kind of eqns?

 

[mitochondria]

Thanks, your answer is correct, however, the reason why I couldn't do it, is because I didn't know the eqn, products... Your chemistry is excellant, could you explain how you got k2c2o4 because this is the trickiest bit. How do I find the products of these kind of eqns?

Trying to figure out whether the product is K₂C₂O₂ or not is actually pretty much a guess-work if you have never come across the reaction before.

Having said that, however, does not mean that you have to do the experiment to predict how the reaction will go.

Upon reading the question, the reactants "oxalic acid" and "potassium hydroxide" should immediately suggest that this reaction is an acid-base reaction (and it follows that you would start thinking about salt and water, which is what happens for many acid-base reactions).

How, if you have never come across oxalic acid before it may look kind of funny to you. Although if you look at the formula (COOH)₂ (don't worry about the H₂O bit), it looks just like the combination of two carboxilic groups and the H just falls off when you react it with a base! (you should have seen carboxilic acids somewhere before, it's a class of organic acid)

So now you can treat oxalic acid as a diprotic acid and write equations with it (provdided that you know how to write equations for simple acid-base reactions)

The salts of oxalic acid are called oxalate.

Hope that helps