3.3 mM of phosphoric acid (H3PO4) is added to water. What will be the pH of the

Question

3.3 mM of phosphoric acid (H3PO4) is added to water. What will be the pH of the sample?
Use the following atomic weights: H:1, O:16, and P:31 g/mol.

Explanation / Answer

H3PO4 HPO4- + H+ Molarity . . . . . . H3PO4 + H2O ==> H3O+ + H2PO4- Initial . . . . . . . . .0.098 . . . . . . . . . . . .0 . . . . . . .0 Change . . . . . . . .-x . . . . . . . . . . . . . .x . . . . . . .x Equilibrium . . . .0.033-x . . . . . . . . . . . x . . . . . . .x Ka1 = [H3O+][H2PO4-] / [H3PO4] = (x)(x) / (0.033-x) = 7.5 x 10^-3 x^2 / (0.033-x) = 7.5 x 10^-3 . . . x^2 / (0.033-x) = 0.0075 x^2 = (0.0075)(0.033-x) x^2 = -0.0075x + 0.000248 x^2 + 0.0075x - 0.000248 = 0.. . . . using the quadratic formula x = 0.00.0124 M = [H3O+] pH = -log [H3O+] = -log (0.0124) = 1.9066 ans

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