Consider 25.0 mL of 0.100M HA (HA is a fictional acid), for which Ka=1.00x10^-5, titrated with 0.0500M NaOH solution. What is the initial pH before adding any titrate?
................HA ===> H^+ + A^-
I...............0.100........0........0
C................-x.............x........x
E...........0.100-x..........x........x
Ka = (H^+)(A^-)/(HA) = 1E-5
Substitute the E line into the Ka expression and solve for x = (H^+), then convert to pH. pH = -log (H^+)
Post your work if you get stuck.
I know how to do the pH afterwards but what's confusing me is that sub into the E line part? Like;
1.00x10^-5 = (H^+) (what goes here though) / (what value goes here?)
How can I find the other values for A- and HA to fill in so I can solve for H+?
To determine the initial pH of the solution before adding any titrant, we need to find the concentration of H+ ions in the acidic solution.
Step 1: Calculate the initial moles of HA:
moles of HA = concentration of HA x volume of HA
moles of HA = 0.100 M x 25.0 mL
moles of HA = 0.100 M x 0.0250 L (since 1 mL = 0.001 L)
moles of HA = 0.0025 mol
Step 2: Since HA is a weak acid, it partially dissociates in water. The equation for its dissociation is:
HA ⇌ H+ + A-
Given the Ka (acid dissociation constant) value of the fictional acid HA as 1.00 x 10^-5, we can set up the equilibrium expression:
Ka = [H+][A-]/[HA]
Step 3: Assume that the concentration of H+ ions at equilibrium is x mol/L. At equilibrium, the concentration of HA will be (0.0025 - x) mol/L, and the concentration of A- will be x mol/L.
Step 4: Substitute the given values into the equilibrium expression and solve for x:
1.00 x 10^-5 = (x)(x) / (0.0025 - x)
Since the value of x is very small compared to 0.0025, we can approximate (0.0025 - x) as 0.0025:
1.00 x 10^-5 = (x)(x) / (0.0025)
Step 5: Rearrange the equation and solve for x:
x^2 = 1.00 x 10^-5 x 0.0025
x^2 = 2.50 x 10^-8
x = √(2.50 x 10^-8)
x = 5.00 x 10^-5
Step 6: Calculate the pH:
pH = -log[H+]
pH = -log(5.00 x 10^-5)
pH ≈ 4.30
Therefore, the initial pH of the solution before adding any titrant is approximately 4.30.
To find the initial pH of the solution before adding any titrant, we need to calculate the concentration of H+ ions in the solution.
Given:
- Initial volume of HA solution (VHA) = 25.0 mL = 0.0250 L
- Concentration of HA solution (CHA) = 0.100 M
- Ka = 1.00x10^-5
To calculate the initial concentration of H+ ions, we can assume that all of the HA dissociates into H+ and A- ions, and we need to find the concentration of H+ ions.
HA ⇌ H+ + A-
The equilibrium expression for Ka can be written as:
Ka = [H+][A-]/[HA]
Since the concentration of A- ions is negligible compared to HA, we can approximate that [A-] ≈ 0. Therefore, we have:
Ka = [H+]/[HA]
We know that [HA] = CHA = 0.100 M and Ka = 1.00x10^-5. Substituting these values into the equation:
1.00x10^-5 = [H+]/0.100
Solving for [H+]:
[H+] = 1.00x10^-5 x 0.100 = 1.00x10^-6 M
Now that we have the initial concentration of H+ ions, we can find the pH using the equation:
pH = -log[H+]
Substituting the value of [H+] into the equation:
pH = -log(1.00x10^-6) ≈ 6.0
Therefore, the initial pH of the solution before adding any titrant is approximately 6.0.