You wish to prepare 0.18 M HNO3 from a stock solution of nitric acid that is 18.9 M. How many milliliters of the stock solution do you require to make up 1.00 L of 0.18 M HNO3?
Answer and.....
HOW DO I DO IT?
To calculate the volume of the stock solution needed to prepare 1.00 L of 0.18 M HNO3, you can use the formula:
C1V1 = C2V2
Where:
C1 = concentration of the stock solution (18.9 M)
V1 = volume of the stock solution needed
C2 = desired concentration (0.18 M)
V2 = final volume (1.00 L)
Now let's plug in the values and solve for V1:
(18.9 M)(V1) = (0.18 M)(1.00 L)
Divide both sides of the equation by 18.9 M:
V1 = (0.18 M)(1.00 L) / 18.9 M
V1 = 0.00952 L
Since we need the answer in milliliters, we can convert liters to milliliters by multiplying by 1000:
V1 = 0.00952 L x 1000 = 9.52 mL
Therefore, you need approximately 9.52 mL of the stock solution to make up 1.00 L of 0.18 M HNO3.
To calculate the volume of the stock solution required to prepare the desired concentration of HNO3, you can use the formula:
C1V1 = C2V2
Where:
C1 = concentration of the stock solution
V1 = volume of the stock solution
C2 = desired concentration of HNO3
V2 = desired final volume of HNO3
In this case, the stock solution concentration (C1) is 18.9 M, and the desired concentration (C2) is 0.18 M. The desired final volume (V2) is 1.00 L.
Using the formula above, we can solve for V1:
(18.9 M)(V1) = (0.18 M)(1.00 L)
Dividing both sides of the equation by 18.9 M:
V1 = (0.18 M)(1.00 L) / 18.9 M
V1 ≈ 0.00952 L
To convert the volume from liters to milliliters, we need to multiply by 1000:
V1 ≈ 0.00952 L * 1000 mL/L
V1 ≈ 9.52 mL
Therefore, you would require approximately 9.52 mL of the 18.9 M stock solution to make up 1.00 L of 0.18 M HNO3.
I gave you the equation last night and filled in the numbers. All you had to do was to solve for the one unknown, mL of the stock solution. What's the problem?
There are a few ways to do these and you may well have a standard formula. However, you might like to try this approach for these problems.
The 1.00 L of 0.18 M HNO3 contains
1.00 L x 0.18 mole L^-1 =
0.18 moles of HNO3. So we need 0.18 moles of HNO3 from the stock solution.
so what volume of 18.9 M HNO3 contains
this number of moles?
volume = 0.18 mole/(18.9 mole L^-1)
=0.009524 L
or 9.5 ml to 2 sig figs