Consider the acid-base equation HF(aq) + CN‾(aq) → (equilibrates with) F‾(aq) + HCN(aq). The Ka of...

Consider the acid-base equation HF(aq) + CN‾(aq) → (equilibrates with) F‾(aq) + HCN(aq).

The K_a of HF is 7.1x10^-4, while the K_b of CN‾ is 2.0x10^-5.

(a) Label each reactant and product as stronger acid, stronger base, weaker acid, or weaker base

(b) Write out the K_a chemical equations for HF and for HCN.

(c) Calculate K_c for the reaction HF(aq) + CN‾(aq) → F‾(aq) + HCN(aq)

Expert Solution

a) weaker acid is HF stromger acid is HCN.

weaker base is CN- stronger base is F-.

b) Ka chemical equation for HF is equal to HF(aq)-> H+(aq) + F-(aq) and Ka equals [H+][F-]/[HF]

Ka chemical equation for HCN is equal toHCN(aq)-> H+(aq) + CN-(aq) and Ka equals [H+][CN-]/[HCN]

c) Kc equals to 1/Ka X 1/Kb

1/7.1*10^-4 X 1/2.0*10^-5

1.4*10³X 5*10⁴

equals 7*10⁷

Hence, K_c equals to 7*10⁷

queen_honey_blossom answered 2 years ago

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