The degree to which a weak base dissociates is given by the base-ionization constant, kb. for the generic weak base, b b(aq)+h2o(l)⇌bh+(aq)+oh−(aq) this constant is given by kb=[bh+][oh−][b] strong bases will have a higher kb value. similarly, strong bases will have a higher percent ionization value. percent ionization=[oh−] equilibrium[b] initial×100% strong bases, for which kb is very large, ionize completely (100%). for weak bases, the percent ionization changes with concentration. the more dilute the solution, the greater the percent ionization. ammonia, nh3, is a weak base with a kb value of 1.8×10−5.parta what is the ph of a 0.245 m ammonia solution? partb what is the percent ionization of ammonia at this concentration?

pH of a 0,245 M ammonia solution is 11,3 and percent ionization is 0,86%

Explanation:

For the equilibrium buffer of NH₃:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻; kb = 1,8x10⁻⁵

kb = [NH₄⁺] [OH⁻] / [NH₃] (1)

When 0,245 M of NH₃ is added, the equilibrium concentrations are:

[NH₃] = 0,245 - x

[NH₄⁺] = x

[OH⁻] = x.

Replacing this values in (1)

x² + 1,8x10⁻⁵x - 4,41x10⁻⁶ = 0

Solving for x:

x = -0,00211 No physical sense. There are not negative concentrations.

x = 0,00211 Real answer

Thus [OH⁻] in equilibrium is 0,00211 M.

As pOH = -log [OH⁻] and 14 = pH + pOH

pH of 0,00211 M is 11,3

It is possible to calculate the percent ionization thus:

Percent ionization = [OH−] equilibrium / [B] initial×100%

Replacing:

0,00211 / 0,245 × 100 = 0,86%

I hope it helps!

ill try to . no promises tho.

explanation:

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.