modibis/vi c d Praine View A&M; Universit..* tion 19 cf 30 eneral Chemistry 4th

Question

modibis/vi c d Praine View A&M; Universit..* tion 19 cf 30 eneral Chemistry 4th Edition University Science Books presanted by Saping Laarninga Phosphoric acid, HsPO4(aq), is a triprotic acid, meaning that one molecule of the acid has three acidic the concentrations of all species in a 6.450 M phosphoric acid solution. 2.16 721 12.32 Estimate the pH, and the concentrations o osplt p soidon 216 721 12.32 Number Number Number Number Numbor pH- D Number Check Answer Next O Previous Previous _up & View

Explanation / Answer

pka = - log (ka)   

ka1 = 6.91 * 10^-3
ka2 = 6.16 * 10^-8
ka3 = 4.78 * 10^-13

3 equilibria of dissociation of H3PO4 is
H3PO4 [H+] + [H2PO4-]
[H2PO4-] [H+] + [HPO42-]
[HPO42-] [H+] + [PO43-]

Where ka1, ka2 and ka3 are the respective equilibrium constants.

H3PO4 [H+] + [H2PO4-]

Initial 0.450M 0 0

at equilibrium ( 0.450M - 0.450x)   0.450x 0.450 x

ie) Ka1 = ( 0.450x)2 /  0.450 ( 1- x)

6.91 * 10^-3 = 0.450x2 /(1-x)

0.00691 -0.00691x = 0.450x2

0.450x2 + 0.00691x -0.00691 = 0

Solving above quadratic equation

x = 0.1165

[H+] = .45 x 0.1165 = 0.05245 M

Thus concentration of [H+] is 5.2 * 10^-2.
This implies that pH is 1.2840 = 1.3

pOH = 12.7

[OH-] = 1.995 x 10^-13

[H2PO4-] is same as [H+]

[H3PO4] = 0.45 - 5.2 * 10^-2 = 0.398 M.

Similarly by solving quadratic equations we get

[HPO42-] = ka2 = 6.16 * 10^-8 M.

[PO4^3-] = 6.69 * 10^-19 M.
The addition of [H+] due to the second and third equilibrium is very low, and so we can neglect it.

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.