Lactic acid, C3H6O3, is found in sour milk. A solution containing 1.00g NaC3H5O3 in 100.0mL of 0.0500M C3H6O3 has a pH= 4.11.
Calculate the Ka of lactic acid
Let lactic acid = HL
The salt, then, is NaL.
I would use the Henderson-Hasselbalch equation but it can be done without that.
Substitute 4.11 for pH.
(base) in the HH equation = 1g/molar mass NaL.
(acid) in the HH equation = 0.05
Solve for pKa and convert to Ka.
To calculate Ka, we need to know the concentration of the conjugate base ([C3H6O3-]) and the concentration of the acid ([HC3H6O3]).
From the information given, we know that the concentration of lactic acid (HC3H6O3) is 0.0500 M.
Next, we need to calculate the concentration of the conjugate base (C3H6O3-). In the solution, there is a dissociation of NaC3H5O3:
NaC3H5O3 → Na+ + C3H5O3-
Since the initial concentration of NaC3H5O3 is 1.00 g in 100.0 mL of solution, we need to convert this to molarity:
1.00 g NaC3H5O3 = 0.0100 mol NaC3H5O3 (by using the molar mass of NaC3H5O3)
100.0 mL = 0.100 L (by converting mL to L)
Now, we can calculate the concentration of the conjugate base:
[C3H6O3-] = 0.0100 mol NaC3H5O3 / 0.100 L = 0.100 M
Next, we use the pH value to find the concentration of H+ ions:
pH = -log[H+]
4.11 = -log[H+]
[H+] = 10^(-4.11)
Since lactic acid is a weak acid, we can assume that most of it dissociates. Therefore, the concentration of H+ is equal to the concentration of the acid:
[HC3H6O3] = [H+] = 10^(-4.11)
Now, we can calculate the Ka using the equation:
Ka = [C3H6O3-] * [H+] / [HC3H6O3]
Plugging in the values we have calculated:
Ka = (0.100 M) * (10^(-4.11)) / (0.0500 M)
Ka = 0.0001
Therefore, the Ka of lactic acid is 0.0001.
To calculate the Ka of lactic acid (C3H6O3), we need to use the given information which includes the amount of lactic acid (C3H6O3) and the pH of the solution. Here's how we can proceed:
Step 1: Write the balanced chemical equation for the dissociation of lactic acid (C3H6O3) in water.
C3H6O3 (aq) ⇌ C3H5O3- (aq) + H+ (aq)
Step 2: Set up the equilibrium expression for the dissociation of lactic acid.
Ka = [C3H5O3-][H+] / [C3H6O3]
Step 3: Rearrange the equilibrium expression to solve for [H+].
[H+] = Ka * [C3H6O3] / [C3H5O3-]
Step 4: Use the pH value to determine [H+].
pH = -log10 [H+]
Step 5: Substitute the known values into the equation and solve for [H+].
[H+] = 10^(-pH)
Step 6: Substitute [H+] into the rearranged equilibrium expression from step 3 and solve for [C3H5O3-].
[C3H5O3-] = [C3H6O3] * [H+] / Ka
Step 7: Substitute the known values into the equation from step 6 and solve for [C3H5O3-].
Step 8: Use the concentration of [C3H5O3-] from step 7 along with the initial concentration of C3H6O3 to solve for Ka.
Ka = [C3H5O3-] * [H+] / [C3H6O3]
Substituting the given values:
Initial concentration of C3H6O3 = 0.0500 M
[C3H5O3-] = amount of NaC3H5O3 / volume of solution = 1.00 g / 0.100 L = 10.0 M (Note: It’s important to convert grams to moles and liters, and divide by 1000 to convert to Molarity)
pH = 4.11
[H+] = 10^(-4.11)
Now substitute these values into the equation from step 8 and solve for Ka:
Ka = (10.0 M) * (10^(-4.11)) / (0.0500 M)
Ka = 7.94 x 10^(-4)
Therefore, the Ka of lactic acid is approximately 7.94 x 10^(-4).