If the value of K_b for NH_3 is 1.8 times 10^-5, calculate the equilibrium const

Question

If the value of K_b for NH_3 is 1.8 times 10^-5, calculate the equilibrium constant for NH_4^+(aq) + H_2O(l) rightarrow NH_3(aq) + H_30^+(aq) 1.8 times 10^-12 5.6 times 10^-10 5.6 times 10^-10 -1.8 times 10^-5 1.8 times 10^-5 The pH of 1.0 M HCOOH(aq) is 1.87. What is the percent ionization of HCOOH? 1.3% 1.8% 1.9% 94% 0% The pH of 0.10 M CH_3COOH(aq) is 2.87. What is the percent ionization of CH_3COOH? 5.0% 1.3% 10% 13% 0.13% What is the pH of 0.24 M HClO(aq) (pk_a = 7.52)? 4.07 3.45 3.76 8.14 0.62 A 0.0010M aqueous solution of a weak acid HX, has a pH of 4.00. What is the percentage dissociation of HX in the solution? 0.0010% 0.010% 0.10% 1.0% 10.% A weak monprotic acid is 0.50% ionized in a 1 0M solution. The dissociation constant, Ka, of the weak acid is 5.0 times 10^-1 5.0 times 10^-3 2.5 times 10^-5 2.5 times 10^-6 2.5 times 10^-7

Explanation / Answer

30.

Kw = Ka*Kb

Kb = 1.8*10^-5

and Kw = 10^-14 always

so

Ka = [NH4][OH-]/[NH4+]

for this

Ka = Kw/Kb = (10^-14)/(1.8*10^-5) = 5.55*10^-10

then choose

C

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