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Phosphoric acid, h3po4(aq, is a triprotic acid, meaning that one molecule of the acid has three acidic protons. estimate the ph, and the concentrations of all species in a 0.300 m phosphoric acid solution. pka1: 2.16 pka2: 7.21 pka3: 12.32 hint: convert the pka values to ka values.

Respuesta :

ka1 = 6.91 * 10^-3
ka2 = 6.16 * 10^-8
ka3 = 4.78 * 10^-13
You can get these by formula: pka = - log (ka)
write 3 equilibria of dissociation of H3PO4
Where ka1, ka2, and ka3 are the respective equilibrium constants.
In the first one, we cannot ignore the decreasing concentration of [H3PO4] and we have a quadratic for the value of concentration of [H+] and thus concentration of [H+] is 4.4 * 10^-2.
This implies that pH is 1.3566. [H2PO4-] is same as [H+] and [H3PO4] = 0.4 - 4.4 * 10^-2 = 0.356.
 

The PH and concentrations of all species are;

pH = 1.31

pH = 1.31[OH-] = 2.06 × 10^(-13) M

pH = 1.31[OH-] = 2.06 × 10^(-13) M[H3PO4] = 0.2514 M

We are given;

pka1: 2.16

pka2: 7.21

pka3: 12.32

Concentration of all species; M = 0.3 M

H3PO4- = 4.59 x 10^(-2) M

HPO2^(4-) = 6.17 x 10^(-8) M

PO3^(4-) = 6.44 x 10^(-19) M

H+ = 4.59 x 10^(-2) M

If we access the dissociation of H3PO4, we will have;

H3PO4 = H+ + (H2PO4)^(2-) Ka = 4.59 x 10^(-2) M

Thus;

4.59 x 10^(-2) = [H+][H2PO42-] / [H3PO4] = (x • x)/(0.1 - x)

Thus;

4.59 x 10^(-2) = x²/(0.1 - x)

Thus;

x² = 0.0459(0.1 - x)

x² = 0.00459 - 0.0459x

x² + 0.0459x - 0.00459 = 0

Thus,

x = 0.0486 M = [H+] = [H2PO42-]

Now,

pH = -log[H+]

pH = -log[0.0486]

pH = 1.31

[OH-] = (1 × 10^(-14))/0.0486

[OH-] = 2.06 × 10^(-13) M

[H3PO4] = 0.3 M - 0.0486 M

[H3PO4] = 0.2514 M

Read more about equilibrium constants at; https://brainly.com/question/3139163

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