What volume of 16 M nitric acid must be diluted with distilled water to prepare 500.0 mL of 0.50 M HNO3?
To find the volume of 16 M nitric acid that must be diluted with water to prepare 500.0 mL of 0.50 M HNO3, we can use the formula:
M1 V1 = M2 V2
Where:
M1 = Initial molarity of the nitric acid solution
V1 = Initial volume of the nitric acid solution
M2 = Final molarity of the desired solution
V2 = Final volume of the desired solution
In this case, we have the following values:
M1 = 16 M (molarity of the initial nitric acid solution)
V1 = Unknown (initial volume of the nitric acid solution)
M2 = 0.50 M (desired final molarity of the solution)
V2 = 500.0 mL (final volume of the desired solution)
Rearranging the formula, we have:
V1 = (M2 * V2) / M1
Substituting the given values, we get:
V1 = (0.50 M * 500.0 mL) / 16 M
V1 = 15.625 mL
Therefore, you would need to dilute 15.625 mL of 16 M nitric acid with distilled water to prepare 500.0 mL of 0.50 M HNO3.
To determine the volume of 16 M nitric acid needed for the dilution, we can use the formula:
C1V1 = C2V2
where:
C1 = initial concentration of the acid (16 M)
V1 = initial volume of the acid (unknown)
C2 = final concentration of the acid (0.50 M)
V2 = final volume of the acid (500.0 mL or 0.500 L)
Let's substitute the values into the formula and solve for V1:
(16 M)(V1) = (0.50 M)(0.500 L)
Dividing both sides of the equation by 16 M, we get:
V1 = (0.50 M)(0.500 L) / 16 M
V1 = 0.015625 L or 15.625 mL
Therefore, you would need to dilute 15.625 mL of 16 M nitric acid with distilled water to prepare 500.0 mL of 0.50 M HNO3.
500 ml of the diluted 0.50 M nitric acid will contain 0.25 moles of HNO3.
1/64 liter of 16M nitric acid contains 16/64 = 0.25 moles of HNO3. So 1/64 liter (62.5 mL) is the answer.