A solution is prepared to be 2.50 x 10^-2 M hypophosphorous acid (H3PO2; Ka = 5.

Question

A solution is prepared to be 2.50 x 10^-2 M hypophosphorous acid (H3PO2; Ka = 5.0 x 10^2) and also 8.50 x 10^-3 M in hypophosphorite ion, H2PO2 .

(a) (0.4 pt)Calculate the pH of the resulting buffer assuming "what you put in is what you get".

(b) (1 pt) Calculate the pH taking into account the "not-so-weak" acid used in the buffer.

**Part (a) I calculated 0.832 to be the pH, I am just not sure about part (b), I know it will be different because the acid being so strong will change how the reaction occurs in some way, I just don't know how to calculate it. Thank you for your help!!

Explanation / Answer

(a)

Ka = 5*10^-2, pKa = 1.30

pH = pKa + log[salt/acid]

      = 1.30 + log [0.0085/0.025] = 0.831

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(b)

H3PO2 < ==> H+ + HPO2^-

Ka = x*x/(0.025-x)

or, 0.05 = x^2/(0.025-x )

or, 0.00125-0.05x -x^2 = 0

or, x = 0.0183M

[H+] = 0.0183 M

pH = -log[0.0183] =1.74

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