The pH of a 0.10 M solution of a certain acid is 3.25 at 25°C. What is the pH of a 0.010 M solution of the same acid at the same temperature?
pH = -log(H^+)
-3.25 = log(H^+)
Solve for (H^+). I estimated 5E-4 but you need to be more accurate than that.
..........HA ==> H^+ + A^-
I.........0.1.....0.....0
C.........-x.....x.....x
E......-0.1-x....x......x
and you know x = 5E-4 from above (again that is an estimate)
Ka = (H^+)(A^-)/(HA)
Solve for Ka.
Then .....HA ==> H^++ A^-
I........0.01.....0....0
C.........-x.....x......x
E......0.01-x.....x.....x
Substitute the E line into Ka expression along with Ka determined from the first part and solve for x = (H^+), then convert to pH.
Oh I see what I did wrong! I got the answer 3.75. Thanks for the help!
To determine the pH of a 0.010 M solution of the same acid at the same temperature, we can use the concept of pH and the properties of dilutions.
The pH of a solution is a measure of its acidity or alkalinity and is calculated using the formula:
pH = -log[H+],
where [H+] represents the concentration of hydrogen ions in moles per liter (M).
In this case, we are given the pH of a 0.10 M solution, which is 3.25. However, we need to find the pH of a more dilute solution with a concentration of 0.010 M.
When a solution is diluted, the moles of solute remain the same, but the volume of the solution increases. This means that the concentration of hydrogen ions, [H+], decreases.
To find the pH of the 0.010 M solution, we can use the relationship between the concentrations and the change in pH caused by dilution. The relationship is as follows:
pH1 - pH2 = log (C1 / C2),
where pH1 and pH2 are the initial and final pH values, and C1 and C2 are the initial and final concentrations of the acid, respectively.
Plugging in the given values, we get:
3.25 - pH2 = log (0.10 / 0.010).
To solve for the pH2, we need to rearrange the equation:
pH2 = 3.25 - log (0.10 / 0.010).
Using logarithmic properties, we can simplify the equation further:
pH2 = 3.25 - log (10).
Since log(10) is equal to 1, the equation becomes:
pH2 = 3.25 - 1.
Finally, we can calculate the pH2:
pH2 = 2.25.
Therefore, the pH of a 0.010 M solution of the same acid at 25°C is 2.25.