Incorrect. Point (e) is the second equivalence point, where the acid has been co

Question

Incorrect. Point (e) is the second equivalence point, where the acid has been completely converted to HPO_3^2-, HPO_3^2- is a weak base with K_b1 = K_w/K_aZ. The concentration of HPO_3^2- must be adjusted for the increase in volume to a total of 50.0 + 100.0 = 150.0 mL, M_2 = M_1 V_1/V_2 = (2.4 M)(50.0 mL)(150.0 mL) As usual, [OH^-] from a weak base can be calculated via K-b = x^2/.80 M - x TildeTilde x^2/0.80 M where x = [OH^-] Since K_b1 >> K_b2 you only need to consider the first ionization.

Explanation / Answer

HPO3-2 + H2O <--> H2PO3- + OH-

Kb = [H2PO3-][OH-]/[HPO3-2]

pKa2 = 7.21 (note that If this is not the case, simply substitute in bolded number)

Kb Kw/Ka = (10^-14)/(10^-7.21) = 1.621*10^-7

so

Kb = [H2PO3-][OH-]/[HPO3-2]

1.621*10^-7 = x*x/(0.80-x)

1.621*10^-7 = x*x/(0.80)

x = sqrt(0.8*1.621*10^-7) = 3.601110*10^-4

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