Editing Weak base (section)

==A typical pH problem==
Calculate the pH and percentage protonation of a .20 M aqueous solution of pyridine, C<sub>5</sub>H<sub>5</sub>N. The K<sub>b</sub> for C<sub>5</sub>H<sub>5</sub>N is 1.8 x 10<sup>−9</sup>.<ref>{{cite web|url=http://www.kentchemistry.com/links/AcidsBases/pHWeakBases.htm|title=Calculations of weak bases|publisher=Mr Kent's Chemistry Page|access-date=2018-03-23}}</ref>

First, write the proton transfer equilibrium:

:<math>\mathrm{H_2O(l) + C_5H_5N(aq) \leftrightarrow C_5H_5NH^+ (aq) + OH^- (aq)}</math>

:<math>K_b=\mathrm{[C_5H_5NH^+] [OH^-]\over [C_5H_5N]}</math>

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

{| width:75%; height:200px border="1"
|+
|-style="height:40px"
! !! C<sub>5</sub>H<sub>5</sub>N !! C<sub>5</sub>H<sub>6</sub>N<sup>+</sup> !! OH<sup>−</sup>
|-
! initial normality
| .20 || 0 || 0
|-
! change in normality
| -x || +x || +x
|-
! equilibrium normality
| .20 -x || x || x
|}

{| width:75%; height:200px border="1"
|-
| Substitute the equilibrium molarities into the basicity constant
| <math>K_b=\mathrm {1.8 \times 10^{-9}} = {x \times x \over .20-x}</math>
|-
| We can assume that x is so small that it will be meaningless by the time we use significant figures.
| <math>\mathrm {1.8 \times 10^{-9}} \approx {x^2 \over .20}</math>
|-
| Solve for x.
| <math>\mathrm x \approx \sqrt{.20 \times (1.8 \times 10^{-9})} = 1.9 \times 10^{-5}</math>
|-
| Check the assumption that x << .20
| <math>\mathrm 1.9 \times 10^{-5} \ll .20</math>; so the approximation is valid
|-
| Find pOH from pOH = -log [OH<sup>−</sup>] with [OH<sup>−</sup>]=x
| <math>\mathrm pOH \approx -log(1.9 \times 10^{-5}) = 4.7 </math>
|-
| From pH = pK<sub>w</sub> - pOH,
| <math>\mathrm pH \approx 14.00 - 4.7 = 9.3</math>
|-
| From the equation for percentage protonated with [HB<sup>+</sup>] = x and [B]<sub>initial</sub> = .20,
| <math>\mathrm percentage \ protonated = {1.9 \times 10^{-5} \over .20} \times 100\% = .0095\% </math>
|}

This means .0095% of the pyridine is in the protonated form of C<sub>5</sub>H<sub>5</sub>NH<sup>+</sup>.