# Weak base

On dissolving in water, a weak base does not dissociate completely and the resulting aqueous solution contains OH- ion and the concerned basic radical in a small proportion along with a large proportion of undissociated molecules of the base.

## pH, Kb, and Kw

Bases range from a pH of greater than 7 (7 is neutral, like pure water) to 14 (though some bases are greater than 14). pH has the formula:

${\displaystyle {\mbox{pH}}=-\log _{10}\left[{\mbox{H}}^{+}\right]}$

Since bases are proton acceptors, the base receives a hydrogen ion from water, H2O, and the remaining H+ concentration in the solution determines pH. 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 results. However, pH of bases is usually calculated using the OH concentration to find the pOH first. This is done because the H+ concentration is not a part of the reaction, while the OH concentration is.

${\displaystyle {\mbox{pOH}}=-\log _{10}\left[{\mbox{OH}}^{-}\right]}$

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

${\displaystyle K_{a}\times K_{b}={[H_{3}O^{+}][NH_{3}] \over [NH_{4}^{+}]}\times {[NH_{4}^{+}][OH^{-}] \over [NH_{3}]}=[H_{3}O^{+}][OH^{-}]}$

Since ${\displaystyle {K_{w}}=[H_{3}O^{+}][OH^{-}]}$ then, ${\displaystyle K_{a}\times K_{b}=K_{w}}$

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

${\displaystyle logK_{a}+logK_{b}=logK_{w}}$

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

${\displaystyle pK_{a}+pK_{b}=pK_{w}=14.00}$

After acquiring pOH from the previous pOH formula, pH can be calculated using the formula pH = pKw - pOH where pKw = 14.00.

Weak bases exist in chemical equilibrium much in the same way as weak acids do, with a base dissociation constant (Kb) indicating the strength of the base. For example, when ammonia is put in water, the following equilibrium is set up:

${\displaystyle \mathrm {K_{b}={[NH_{4}^{+}][OH^{-}] \over [NH_{3}]}} }$

Bases that have a large Kb will ionize more completely and are thus stronger bases. As stated above, pH of the solution depends on the H+ concentration, which is related to the OH concentration by the self-ionization constant (Kw = 1.0x10−14). 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.[1]

## Percentage protonated

As seen above, the strength of a base depends primarily on pH. 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 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.[2]

The typical proton transfer equilibrium appears as such:

${\displaystyle B(aq)+H_{2}O(l)\leftrightarrow HB^{+}(aq)+OH^{-}(aq)}$

B represents the base.

${\displaystyle Percentage\ protonated={molarity\ of\ HB^{+} \over \ initial\ molarity\ of\ B}\times 100\%={[{HB}^{+}] \over [B]_{initial}}{\times 100\%}}$

In this formula, [B]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.[3]

First, write the proton transfer equilibrium:

${\displaystyle \mathrm {H_{2}O(l)+C_{5}H_{5}N(aq)\leftrightarrow C_{5}H_{5}NH^{+}(aq)+OH^{-}(aq)} }$
${\displaystyle K_{b}=\mathrm {[C_{5}H_{5}NH^{+}][OH^{-}] \over [C_{5}H_{5}N]} }$

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

C5H5N C5H6N+ OH
initial normality .20 0 0
change in normality -x +x +x
equilibrium normality .20 -x x x
 Substitute the equilibrium molarities into the basicity constant ${\displaystyle K_{b}=\mathrm {1.8\times 10^{-9}} ={x\times x \over .20-x}}$ We can assume that x is so small that it will be meaningless by the time we use significant figures. ${\displaystyle \mathrm {1.8\times 10^{-9}} \approx {x^{2} \over .20}}$ Solve for x. ${\displaystyle \mathrm {x} \approx {\sqrt {.20\times (1.8\times 10^{-9})}}=1.9\times 10^{-5}}$ Check the assumption that x << .20 ${\displaystyle \mathrm {1} .9\times 10^{-5}\ll .20}$; so the approximation is valid Find pOH from pOH = -log [OH−] with [OH−]=x ${\displaystyle \mathrm {p} OH\approx -log(1.9\times 10^{-5})=4.7}$ From pH = pKw - pOH, ${\displaystyle \mathrm {p} H\approx 14.00-4.7=9.3}$ From the equation for percentage protonated with [HB+] = x and [B]initial = .20, ${\displaystyle \mathrm {p} ercentage\ protonated={1.9\times 10^{-5} \over .20}\times 100\%=.0095\%}$

This means .0095% of the pyridine is in the protonated form of C5H5NH+.

## Simple Facts

• An example of a weak base is ammonia. It does not contain hydroxide ions, but it reacts with water to produce ammonium ions and hydroxide ions.[4]
• The position of equilibrium varies from base to base when a weak base reacts with water. The further to the left it is, the weaker the base.[5]
• When there is a hydrogen ion gradient between two sides of the biological membrane, the concentration of some weak bases are focused on only one side of the membrane.[6] Weak bases tend to build up in acidic fluids.[6] Acid gastric contains a higher concentration of weak base than plasma.[6] Acid urine, compared to alkaline urine, excretes weak bases at a faster rate.[6]