What is the pH of a solution prepared by diluting 25.00mL of 0.020M barium hydroxide to a volume of 100.00mL?
Start with 0.020 M Ba(OH)2.
Calculate the (OH^-). Ba(OH)2 is 100% ionized.
Then dilute that by a factor of 25/100.
Then POH and pH from that. Post your work if you need more help.
Find the moles of Ba(OH)2 by .025L*.020M
Then divide answer by .100L to find new molarity
then find pOH=-log(OH-)
Then subtract from 14
pH=14-pOH
To find the pH of a solution prepared by dilution, we first need to calculate the concentration of the diluted solution.
In this case, we are diluting 25.00 mL of a 0.020 M solution of barium hydroxide to a volume of 100.00 mL.
We can use the formula for dilution:
C1V1 = C2V2
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
Let's plug in the values we have:
(0.020 M)(25.00 mL) = C2(100.00 mL)
Now let's solve for C2, the final concentration:
C2 = (0.020 M)(25.00 mL) / (100.00 mL)
C2 = 0.005 M
Therefore, the final concentration of the diluted solution is 0.005 M.
To find the pH of this solution, we need to know the nature of the solute (barium hydroxide in this case). If it is a strong base, we can follow a simple method:
For a strong base like barium hydroxide, the pH of the solution can be calculated using the formula:
pOH = -log10([OH-])
Since barium hydroxide dissociates completely in water, the concentration of hydroxide ions ([OH-]) will be the same as the concentration of the diluted solution, which is 0.005 M in this case.
Now, let's calculate the pOH using the formula:
pOH = -log10(0.005)
pOH = 2.30
Finally, to find the pH, we can use the equation:
pH = 14 - pOH
pH = 14 - 2.30
pH ≈ 11.70
Therefore, the pH of the solution prepared by diluting 25.00 mL of 0.020 M barium hydroxide to a volume of 100.00 mL is approximately 11.70.
To find the pH of a solution prepared by diluting a given concentration, we need to use the concept of dilution and the properties of the compound.
First, let's understand the process of dilution. When a solute (in this case, barium hydroxide) is diluted with a solvent (water), the amount of solute remains constant while the volume of the solution increases. This process reduces the concentration of the solute.
In this case, you have 25.00 mL of a 0.020 M (molar) solution of barium hydroxide, which is being diluted to a final volume of 100.00 mL.
To calculate the final concentration, we can use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution
V1 = initial volume of the solution
C2 = final concentration of the solution
V2 = final volume of the solution
Plugging in the values, we get:
(0.020 M)(25.00 mL) = C2(100.00 mL)
Simplifying the equation, we find that the final concentration (C2) is 0.005 M.
Now, to determine the pH of the solution, we need to know the nature of barium hydroxide. Barium hydroxide is a strong base. When it dissolves in water, it dissociates completely into its ions. In this case, barium hydroxide dissociates into one barium ion (Ba2+) and two hydroxide ions (OH-).
Since it is a strong base, we can assume that all of the hydroxide ions will react with water to form hydroxide ions (OH-) and hydronium ions (H3O+). The concentration of hydroxide ions (OH-) in the solution will be equal to the initial concentration of barium hydroxide.
Therefore, the concentration of the hydroxide ions (OH-) in the solution is 0.020 M.
Using the concentration of hydroxide ions, we can calculate the pOH of the solution using the formula:
pOH = -log[OH-]
Plugging in the value, we find that the pOH is -log(0.020) = 1.70.
Finally, to find the pH of the solution, we use the relation:
pH + pOH = 14
Rearranging the equation, we have:
pH = 14 - pOH
Substituting the value of pOH into the equation, we find that the pH of the solution is:
pH = 14 - 1.70 = 12.30.
Therefore, the pH of the solution prepared by diluting 25.00 mL of 0.020 M barium hydroxide to a volume of 100.00 mL is 12.30.