Find the concentration of the polyprotic weak acid given the following:
- 1 equiv needs 6 ml NaOH, 2 equiv needs 18 ml NaOH, 3 equiv needs 28 ml NaOH
- pKa1 = 2, pKa2 = 3.9, pKa3, 10
- Molarity of NaOH = 0.175
- Total volume of NaOH = 28 ml
- Volume of weak acid = 30 ml
Please help! Thanks!
To find the concentration of the polyprotic weak acid, we need to use the information provided and multi-step calculations. Here's how you can do it:
Step 1: Calculate the moles of NaOH used for each equivalence point.
- 1 equiv needs 6 ml = 0.006 L of NaOH, therefore moles of NaOH for 1 equiv = 0.006 L * 0.175 M = 0.00105 mol NaOH
- 2 equiv needs 18 ml = 0.018 L of NaOH, therefore moles of NaOH for 2 equiv = 0.018 L * 0.175 M = 0.00315 mol NaOH
- 3 equiv needs 28 ml = 0.028 L of NaOH, therefore moles of NaOH for 3 equiv = 0.028 L * 0.175 M = 0.0049 mol NaOH
Step 2: Calculate the moles of weak acid used for each equivalence point.
Since the volume of weak acid is 30 ml = 0.03 L, the moles of weak acid for each equivalence point can be calculated using the concentration formula: Moles = Concentration * Volume.
- Moles of weak acid for 1 equiv = x * 0.03 L (where x is the concentration of the weak acid)
- Moles of weak acid for 2 equiv = 2x * 0.03 L
- Moles of weak acid for 3 equiv = 3x * 0.03 L
Step 3: Express the ionization of the weak acid and write down the molecular and ionic equations for the dissociation of the acid.
Since the weak acid is polyprotic, it will undergo successive ionizations. The molecular equation for the first ionization can be written as:
HA <=> H+ + A-
The molecular equation for the second ionization is:
A- <=> H+ + A2-
The molecular equation for the third ionization is:
A2- <=> H+ + A3-
Step 4: Use the Ka expression to calculate the concentration of the weak acid at each equivalence point.
The Ka expression for the first ionization is:
Ka1 = [H+][A-] / [HA]
The Ka expression for the second ionization is:
Ka2 = [H+][A2-] / [A-]
The Ka expression for the third ionization is:
Ka3 = [H+][A3-] / [A2-]
The values of pKa1, pKa2, and pKa3 are given as 2, 3.9, and 10, respectively. We can use the pKa values to calculate the Ka values.
Ka1 = 10^(-pKa1) = 10^(-2) = 0.01
Ka2 = 10^(-pKa2) = 10^(-3.9) = 0.00079
Ka3 = 10^(-pKa3) = 10^(-10) = 1e-10
Step 5: Set up equations using the Ka values and solve for the unknown concentration.
- For the first ionization:
0.01 = ([H+][A-]) / [HA] = (x * 0.03) * (x * 0.03) / (x * 0.03) = (x * 0.03) / (x * 0.03) = 1
- For the second ionization:
0.00079 = ([H+][A2-]) / [A-] = (2x * 0.03) * (2x * 0.03) / (x * 0.03) = (4x^2 * 0.03) / (x * 0.03) = 4x
- For the third ionization:
1e-10 = ([H+][A3-]) / [A2-] = (3x * 0.03) * (3x * 0.03) / (x * 0.03) = (9x^2 * 0.03) / (x * 0.03) = 9x
Simplifying each equation, we get:
1 = x
4x = 0.00079
9x = 1e-10
Solving the second equation, we find x = 0.0001975 M (concentration of weak acid).
Therefore, the concentration of the polyprotic weak acid is 0.0001975 M.