If 1,2,3,4 moles of H3po4, nah2po4, na2hpo4, and na3po4 are mixed in an aqueous solution.find the pH.
Ka1=10^-3, Ka2=10^-7, ka3= 10^-13
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To find the pH of the solution, we need to consider the dissociation of each species and determine the concentration of H+ ions.
First, let's write down the dissociation reactions for the given compounds:
1. H3PO4(aq) ⇌ H+ + H2PO4-
2. NaH2PO4(aq) ⇌ Na+ + H+ + HPO42-
3. Na2HPO4(aq) ⇌ 2Na+ + H+ + HPO42-
4. Na3PO4(aq) ⇌ 3Na+ + PO43-
We are given the Ka values, which represent the equilibrium constant for each dissociation reaction. These Ka values can be used to determine the degree of dissociation for each compound.
The degree of dissociation, α, measures the fraction of a compound that has dissociated into its respective ions. We can calculate α using the following formula:
α = √(Ka * C)
Where Ka is the dissociation constant and C is the initial concentration of the compound.
Let's calculate the concentration of each compound:
- For H3PO4, the initial concentration is given as 1 mole.
- For NaH2PO4, the initial concentration is given as 2 moles.
- For Na2HPO4, the initial concentration is given as 3 moles.
- For Na3PO4, the initial concentration is given as 4 moles.
Now, let's calculate the concentrations of H+ ions using the dissociation reactions:
1. For H3PO4:
α = √(Ka1 * C) = √(10^-3 * 1) = 0.0316
Since H3PO4 dissociates to give one H+ ion, the concentration of H+ ions, [H+], is equal to the degree of dissociation, α.
[H+] = 0.0316 M
2. For NaH2PO4:
α = √(Ka2 * C) = √(10^-7 * 2) = 0.00447
Since NaH2PO4 dissociates to give one H+ ion, the concentration of H+ ions, [H+], is equal to the degree of dissociation, α.
[H+] = 0.00447 M
3. For Na2HPO4:
α = √(Ka2 * C) = √(10^-7 * 3) = 0.00653
Since Na2HPO4 dissociates to give one H+ ion, the concentration of H+ ions, [H+], is equal to the degree of dissociation, α.
[H+] = 0.00653 M
4. For Na3PO4:
α = √(Ka3 * C) = √(10^-13 * 4) = 0.002
Since Na3PO4 dissociates to give three H+ ions, the concentration of H+ ions, [H+], is three times the degree of dissociation, α.
[H+] = 3 * 0.002 = 0.006 M
Now, let's calculate the total concentration of H+ ions in the solution by summing up the concentrations from each compound:
[H+]total = [H+]H3PO4 + [H+]NaH2PO4 + [H+]Na2HPO4 + [H+]Na3PO4
[H+]total = 0.0316 + 0.00447 + 0.00653 + 0.006
[H+]total = 0.0486 M
Finally, to calculate the pH, we use the equation:
pH = -log[H+]
pH = -log(0.0486)
pH ≈ 1.314
Therefore, the pH of the aqueous solution is approximately 1.314.