Question

Find Ph of Na3PO4 that is 0.5 Molar such that Ka=2.4*10^-2

Answer 1

12.99

Answer 2

Na3PO4 is a salt of a strong base (NaOH) and a weak acid (H3PO4). When dissolved in water, the phosphate ion (PO4^3-) acts as a base and undergoes hydrolysis.

H3PO4 is a triprotic acid with three dissociation constants: Ka1, Ka2, and Ka3. The relevant hydrolysis reaction for PO4^3- is:

PO4^3-(aq) + H2O(l) <=> HPO4^2-(aq) + OH-(aq)

The equilibrium constant for this reaction is Kb1. This Kb1 is related to the third dissociation constant of phosphoric acid (Ka3) by the ion product of water (Kw):

Kb1 = Kw / Ka3

Standard values for phosphoric acid are: Ka1 = 7.1 × 10^-3 Ka2 = 6.3 × 10^-8 Ka3 = 4.2 × 10^-13 Kw = 1.0 × 10^-14 at 25°C.

Using standard values, Kb1 = (1.0 × 10^-14) / (4.2 × 10^-13) = 0.0238 ≈ 2.38 × 10^-2.

The question provides "Ka = 2.4 * 10^-2". This value is remarkably close to the calculated Kb1 (2.38 × 10^-2). Given that PO4^3- is a base, it is highly probable that the provided "Ka" is actually intended to be the Kb for the hydrolysis of PO4^3-, or it is Ka3, but with a value that is very close to Kb1. If it were Ka3, the Kb1 would be Kw/Ka3 = 10^-14 / (2.4 * 10^-2) = 4.167 * 10^-13, leading to a pH near 7, which is incorrect for Na3PO4. Therefore, we will proceed by assuming the given value, 2.4 * 10^-2, is the Kb for the first hydrolysis step of PO4^3-.

Let C be the initial concentration of Na3PO4, so C = 0.5 M. Let x be the concentration of OH- produced at equilibrium.

Initial: [PO4^3-] = 0.5 M, [HPO4^2-] = 0, [OH-] = 0 Change: [PO4^3-] = -x, [HPO4^2-] = +x, [OH-] = +x Equilibrium: [PO4^3-] = 0.5 - x, [HPO4^2-] = x, [OH-] = x

The expression for Kb is: $K_{b1} = \frac{[HPO_4^{2-}][OH^-]}{[PO_4^{3-}]}$

Substituting the equilibrium concentrations and the given Kb value: $2.4 \times 10^{-2} = \frac{x \cdot x}{0.5 - x}$ $x^2 = 2.4 \times 10^{-2} (0.5 - x)$ $x^2 = 0.012 - 0.024x$

Rearranging into a quadratic equation: $x^2 + 0.024x - 0.012 = 0$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-0.024 \pm \sqrt{(0.024)^2 - 4(1)(-0.012)}}{2(1)}$ $x = \frac{-0.024 \pm \sqrt{0.000576 + 0.048}}{2}$ $x = \frac{-0.024 \pm \sqrt{0.048576}}{2}$ $x = \frac{-0.024 \pm 0.220399}{2}$

Since x represents a concentration, it must be positive: $x = \frac{-0.024 + 0.220399}{2}$ $x = \frac{0.196399}{2}$ $x = 0.0981995$ M

So, $[OH^-] = 0.0981995$ M.

Now, calculate pOH: $pOH = -log[OH^-]$ $pOH = -log(0.0981995)$ $pOH \approx -log(9.82 \times 10^{-2})$ $pOH = 2 - log(9.82)$ $pOH \approx 2 - 0.9921$ $pOH \approx 1.0079$

Finally, calculate pH: $pH = 14 - pOH$ $pH = 14 - 1.0079$ $pH = 12.9921$

The pH of the 0.5 M Na3PO4 solution is approximately 12.99.

Explanation of the solution:

1. Identify Na3PO4 as a salt of a strong base and a weak acid (H3PO4). The PO4^3- ion is the conjugate base of HPO4^2-.
2. The dominant reaction is the hydrolysis of PO4^3- producing OH-: PO4^3- + H2O <=> HPO4^2- + OH-.
3. Assume the given "Ka = 2.4 * 10^-2" is actually the Kb for this hydrolysis, as it is very close to the standard Kb1 for PO4^3- and a pH around 13 is expected for a strong base like PO4^3-.
4. Set up an ICE table for the hydrolysis reaction and write the Kb expression.
5. Solve the quadratic equation for [OH-].
6. Calculate pOH from [OH-].
7. Calculate pH from pOH using pH + pOH = 14.