An experiment requires 0.7 M NH3(aq). The
stockroom manager estimates that 15 L of
the base is needed. What volume of 15 M
NH3(aq) will be used to prepare this amount of 0.7 M base?
Answer in units of mL
Use c1v1 = c2v2
c = concn
v = volume
Substitute the numbers and solve for the unknown.
To find the volume of 15 M NH3(aq) needed to prepare 0.7 M NH3(aq), we can use the concept of dilution formula. The formula for dilution is:
C1V1 = C2V2
Where:
C1 = Initial concentration
V1 = Initial volume
C2 = Final concentration
V2 = Final volume
In this case, we know:
C1 = 15 M (concentration of the stock solution)
V1 = ?? (volume of the stock solution to be used)
C2 = 0.7 M (desired final concentration)
V2 = 15 L (desired final volume in liters)
Let's plug in these values into the dilution formula:
(15 M)(V1) = (0.7 M)(15 L)
To solve for V1 (volume of the stock solution), we can rearrange the equation:
V1 = (0.7 M)(15 L) / 15 M
V1 = (0.7 * 15) / 15
V1 = 0.7 L
Since the answer needs to be in milliliters, we can convert it:
0.7 L = 700 mL
Therefore, to prepare 0.7 M NH3(aq), you will need to use 700 mL of 15 M NH3(aq).
To find the volume of 15 M NH3(aq) needed to prepare 0.7 M base, we can use the following formula:
C1V1 = C2V2
Where:
C1 = initial concentration of NH3(aq) = 15 M
V1 = initial volume of NH3(aq) (unknown)
C2 = final concentration of NH3(aq) = 0.7 M
V2 = final volume of NH3(aq) = 15 L
Rearranging the formula, we have:
V1 = (C2 * V2) / C1
Substituting the given values into the formula:
V1 = (0.7 M * 15 L) / 15 M
We can cancel out the units to simplify:
V1 = 0.7 L
Since the answer requires the volume in mL, we can convert liters to milliliters by multiplying by 1000:
V1 = 0.7 L * 1000 mL/L
Performing the calculation:
V1 = 700 mL
Therefore, the volume of 15 M NH3(aq) needed to prepare 0.7 M base is 700 mL.