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In chemistry, a weak base is a Chemical Base that does not Ionize fully in an Aqueous Solution . As bases are proton acceptors, a weak base may also be defined as a chemical base in which Protonation is incomplete. This results in a relatively low PH level compared to Strong Base s. Bases range from a pH of greater than 7 (7 is neutral, like pure water) to 14 (though some bases are greater than 14). The pH level has the formula:
:\mbox{pH} = -\log_{10} \left \mbox{H}^+ ight
Since bases are Proton acceptors, the base receives a hydrogen ion from water, H2O, and the remaining H+ Concentration in the solution determines the pH level. Weak bases will have a higher H+ concentration because they are less completely protonated than stronger bases and, therefore, more hydrogen ions remain in the solution. If you plug in a higher H+ concentration into the formula, a low pH level results. However, the pH level of bases is usually calculated using the OH- concentration to find the pOH level first. This is done because the H+ concentration is not a part of the reaction, while the OH- concentration is.
:\mbox{pOH} = \log_{10} \left \mbox{OH}^- ight

By multiplying a conjugate acid (such as NH4+) and a conjugate base (such as NH3) the following is given:

: K_a imes K_b = {[H_3O^+][NH_3]\over[NH_4^+]} imes {[NH_4^+] = [H_3O^+ [OH^-]

Since {K_w} = {Link without Title} {Link without Title} then, ''K_a imes K_b = K_w''

By taking logarithms of both sides of the equation, the following is reached:

:logK_a + logK_b = logK_w

Finally, multipying throughout the equation by -1, the equation turns into:

:pK_w + pK_b = pK_a = 14.00

After acquiring pOH from the previous pOH formula, pH can be calculated using the formula pH = pKw - pOH where pKw = your mom

Weak bases exist in Chemical Equilibrium much in the same way as Weak Acid s do, with a Base Ionization Constant (Kb) (or the '''Base Dissociation Constant''') indicating the strength of the base. For example, when ammonia is put in water, the following equilibrium is set up:

:\mathrm{K_b={ {Link without Title} {Link without Title} \over {Link without Title} }}

Bases that have a large Kb will ionize more completely and are thus stronger bases. As stated above, the pH of the solution depends on the H+ concentration, which is related to the OH- concentration by the Ionic Constant of water (Kw = 1.0x10-14) (See article Self-ionization Of Water .) A strong base has a lower H+ concentration because they are fully protonated and less hydrogen ions remain in the solution. A lower H+ concentration also means a higher OH- concentration and therefore, a larger Kb.



NaOH (s) (sodium hydroxide) is a stronger base than (CH3CH2)2NH (l) (diethylamine) which is a stronger base than NH3 (g) (ammonia). As the bases get weaker, the smaller the Kb values become. The pie-chart representation is as follows:

  • purple areas represent the fraction of OH- ions formed

  • red areas represent the cation remaining after ionization

  • yellow areas represent dissolved but non-ionized molecules.




PERCENTAGE PROTONATED

As seen above, the strength of a base depends primarily on the pH level. To help describe the strengths of weak bases, it is helpful to know the percentage protonated-the percentage of base molecules that have been protonated. A lower percentage will correspond with a lower pH level because both numbers result from the amount of protonation. A weak base is less protonated, leading to a lower pH and a lower percentage protonated.

The typical proton transfer equilibrium appears as such:

:B(aq) + H_2O(l) \leftrightarrow HB^+(aq) + OH^-(aq)

B represents the base.

:Percentage\ protonated = {molarity\ of\ HB^+ \over\ initial\ molarity\ of\ B} imes 100\% = { [B _{initial}} { imes 100\%}

In this formula, {Link without Title} initial is the initial molar concentration of the base, assuming that no protonation has occurred.


A TYPICAL PH PROBLEM


Calculate the pH and percentage protonation of a .20 M aqueous solution of pyridine, C5H5N. The Kb for C5H5N is 1.8 x 10-9.

First, write the proton transfer equilibrium:

:\mathrm{H_2O(l) + C_5H_5N(aq) \leftrightarrow C_5H_6N^+ (aq) + OH^- (aq)}

:K_b=\mathrm{ [C_5H_5N }

The equilibrium table, with all concentrations in moles per liter, is


  <math>\mathrm X \approx \sqrt{20 Imes (18 Imes 10^{-9})} 19 imes 10^{-5}</math>
  Find POH From POH -log with [OH<sup>-</sup> =x
  <math>\mathrm POH \approx -log(19 Imes 10^{-5}) 47 </math>
  From PH pK<sub>w</sub> - pOH,
  <math>\mathrm PH \approx 1400 - 47 93</math>
  From The Equation For Percentage Protonated With "HB<sup>+</sup>]" class="copylinks" target="_blank">= x and [B <sub>initial</sub> = 20,
  <math>\mathrm Percentage \ Protonated {19 imes 10^{-5} \over 20} imes 100\% = 0095\% </math>