Complete the equations for the following equillibria and calculate Keq where the Keq expression includes [H2O]. Be sure to enter Keq in proper scientific notation.

Species: NH3 pKa:35

Species: NH4+ pKa:9.25

Species: H2O pKa:15.74

Species: H30+ pKa:-1.74

(A) Ammonia (acting as a base) reacts with water (acting as an acid)

NH3 + H2O <—> Keq=?

(B)Ammonia (acting as an acid) reacts with water (acting as an base)

NH3 + H2O <—> Keq=?

(C) Which of these reactions is more important in an aqueous solution of ammonia. A or B?

Having lots of issues on this organic Hw prob. Any help is appreciated thank you!!

3 Answers

  • A) NH3 +H2O <–> NH4+ (+) OH-

    Log(Keq) = pKa [B] – pKa [A] … therefore Keq = 10^ (pKa [B]- pKa [A])

    Keq = 10^ (pKa [B]- pKa [A])

    Keq = 10^ (9.25 – 15.74)

    Keq = 3.2×10^-7

    B) NH3 + H20 <–> NH2- (+) H30+

    Keq = 10^ (-1.74 – 35)

    Keq = 1.8×10^-37

    C) The answer is “A”.

  • I am also in twelfth and i have the same hindrance dude….. However are attempting learning them by writing and practising… Hope it’s going to work…’cos i’m doing the identical too…. Pleasant of good fortune…

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