IF A 50.0 ML SAMPLE OF CALCIUM HYDROXIDE IS TITRATED WITH 25.0 ML OF 0.200M PHOSPHORIC ACID, WHAT IS THE MOLARITY OF THE BASE?
3Ca(OH)2 + 2H3PO4 ==> 6H2O + Ca3(PO4)2
mols H3PO4 = M x L = ?
Using the coefficients in the balanced equation, convert mols H3PO4 to mols Ca(OH)2. That will be
mols H3PO4 x (3 mols Ca(OH)2/2 mols H3PO4) = mols H3PO4 x (3/2) = ?
Then M Ca(OH)2 = mols Ca(OH)2/L Ca(OH)2
To find the molarity of the base (calcium hydroxide), you need to use the concept of stoichiometry and the balanced chemical equation for the reaction between calcium hydroxide and phosphoric acid. The balanced equation is:
3Ca(OH)2 + 2H3PO4 -> Ca3(PO4)2 + 6H2O
Given:
Volume of calcium hydroxide solution (base) = 50.0 mL
Volume of phosphoric acid solution (acid) = 25.0 mL
Molarity of phosphoric acid solution (acid) = 0.200 M
First, determine the number of moles of acid used in the reaction by using the formula:
moles of acid (H3PO4) = Molarity of acid (H3PO4) x Volume of acid (H3PO4)
moles of acid = 0.200 M x 0.0250 L (since 1 mL = 0.001 L)
Next, since the balanced equation shows that 3 moles of calcium hydroxide will react with 2 moles of phosphoric acid, use stoichiometry to determine the number of moles of base (Ca(OH)2) used in the reaction.
moles of base (Ca(OH)2) = (moles of acid (H3PO4) x 3) / 2
Now, calculate the molarity of the base by dividing the moles of base by the volume of the base solution:
Molarity of the base (Ca(OH)2) = moles of base (Ca(OH)2) / Volume of base solution (Ca(OH)2)
Given that the volume of the base solution is 50.0 mL (converted to liters, this is 0.050 L), substitute the values into the equation:
Molarity of the base (Ca(OH)2) = (moles of acid (H3PO4) x 3) / (2 x Volume of base solution (Ca(OH)2))
Now you can calculate the molarity of the base by plugging in the values calculated above.