In aqueous solution, when HA is reacted with Na3B, NaH2B is one of the products. H3B: pKa1 = 4.75, pka2 = 6.80, and pKa3 = 10.45

HA: pKa = 2.25

a. Write the complete molecular, complete ionic, and net ionic equations for the reaction of HA with

Na3B.

CME:

CIE:

NIE:

b. Calculate Kn for the net ionic equation.

It would be c. risk.

Answer:             1.0  × 10⁻⁵ msolution:                     as we know ph is given as,                                    ph  = -log [h⁺]and,                                    [h⁺]  =  10^-ph    (1)putting value of ph in eq. 1,                                    [h⁺]  =  10⁻⁵                                    [h⁺]  =  1.0  × 10⁻⁵ m

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