Chemistry - Acids and Bases

Question

Which one of the following acids require the largest volume of 1.00 mol/dm3 potassium hydoxide o neutralize? (A) 1.00 dm3 0.50 mol/dm3 H3PO4 (B) 1.00 dm3 1.30 mol/dm3 CH3COOH. Answer given is (A) but i don't understand how. Thank you.

Answer 1

There are two methods you can use to solve this problem. I will present one method below.

 

First write a balanced equation:

METHOD 1

 

(A) 3KOH + H3PO4 ----> K3PO4 + 3H20

 

There is a relationship between the molarities of the reactants, volumes (in dm3 or liter), and the mole ratio of reactants. Thus:

 

(Volume of KOH  x Molarity of KOH ) / ( Volume of H3PO4 x Molarity of H3PO4) = (Mole of KOH) / (Mole of H3PO4)

 

The mole ratio is from the balanced equation.  That is, 3 moles of KOH to 1 mole of H3PO4(from the balanced equation)

 

Information given are:

 

Volume of KOH = ?;  Molarity of KOH = 1.00 dm3; Volume of H3PO4 = 1.00 dm3; Molarity of H3PO4 = 0.50 mol/dm3

 

Substituting these figures, we have the following:

 

(Volume of KOH  x 1.00 mol/dm3) / (1.00 dm3  x 0.50 mol/dm3 = 3/1    where  the symbol / stands for the division sign

 

Cross multiplying, we have:

 

Volume of KOH  x 1.00  x  1 = 3x1.00x0.50

 

Volume of KOH  x 1.00 = 1.50

 

Therefore, Volume of KOH required = 1.50/1.00  = 1.50 dm3

 

Using the same method for the neutralization of acetic acid (CH3COOH), you will obtain 1.30 dm3.

 

Therefore (A) is the answer because 1.50 dm3 is greater than 1.30 dm3.

 

Here is the balanced equation for the reaction of KOH with CH3COOH:

 

(B) KOH + CH3COOH -----------> CH3COOK + H2O Here, the mole ratio is 1/1