What volume of 5x10^-3 M HNO3 is needed to titrate 100 mL of 5x10^-3 M Ca(OH)2 to the equivalence point?
(A) 100 mL
(B) 200 mL
I believe the answer is (B), my teacher believes it is (A). This is the difference between a B and an A for me.
-Eric
Since both HNO3 and Ca(OH)2 is a strong reagent, you will not observe a buffer between the two, so the equivalence point will be at a pH of 7, which means that moles of H+ equals moles of OH- at that point.
HNO3 contains 1 mol of H+ because: 1HNO3 =
1H+ plus 1NO3-
Ca(OH)2 has 2 mol of OH- because: 1Ca(OH)2 = 1Ca+2 plus
2OH-
Remember that it is the
moles that matter at equivalence point.
You are adding HNO3, because you are
titrating Ca(OH)2.
0.100 L of 5x10^-3 M Ca(OH)2 will give you 0.0005 moles. But this is only for 1 mol of OH-. So multiply this number by two to give you two moles of OH: 2x0.0005 =
0.001 mol
From this point, all you have to find is the amount of HNO3 needed to give you 0.001 mol (Because remember mol H = mol OH at EQ point).
0.005M HNO3 * 0.2L =
0.001 mol
So you need to add 200 mL of HNO3 to reach the equivalence point.
What if the question asked:
What volume of 5x10^-3 M Ca(OH)2 is needed to titrate 100 mL of 5x10^-3 M HNO3 to the equivalence point?
You are
adding Ca(OH)2 this time.
Now you would do: 5x10^-3 M HNO3 * 0.1 L =
0.0005 mol
So we have to find a mole quantity equal to this number; and remember Ca(OH)2 has two moles of OH. Therefore, we can work backwards:
0.0005 mol / 2 OH = 0.00025 mol
0.00025 mol / 5x10^-3 M Ca(OH)2 = 0.05 L or
50 mL of Ca(OH)2 is needed.
