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my blog post ipad repair hanover park In chemistry, a weak base is a chemical base that does not ionize fully in an aqueous solution. As Brønsted–Lowry 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 compared to strong bases. 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:

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

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.

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

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

$K_a\times K_b=\frac{[H_3{O}^{+}][N{H}_3]}{[N{H}_4^{+}]}\times \frac{[N{H}_4^{+}][O{H}^{-}]}{[N{H}_3]}=[H_3{O}^{+}][O{H}^{-}]$

Since $K_w=[H_3{O}^{+}][O{H}^{-}]$ then, $K_a\times K_b=K_w$

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

$log{K}_a+log{K}_b=log{K}_w$

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

$p{K}_a+p{K}_b=p{K}_w=14.00$

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

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

${\mathrm{K}}_{\mathrm{b}}=\frac{[N{H}_4^{+}][O{H}^{-}]}{[N{H}_3]}$

Bases that have a large K_b 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 (K_w = 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 K_b.

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 K_b values become.

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.

The typical proton transfer equilibrium appears as such:

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

B represents the base.

$Percentageprotonated=\frac{molarityofH{B}^{+}}{initialmolarityofB}\times 100\%=\frac{[{HB}^{+}]}{[B{]}_{initial}}\times 100\%$

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 K_b for C₅H₅N is 1.8 x 10⁻⁹.

First, write the proton transfer equilibrium:

${\mathrm{H}}_{\mathrm{2}}\mathrm{O}\mathrm{(}\mathrm{l}\mathrm{)}+{\mathrm{C}}_{\mathrm{5}}{\mathrm{H}}_{\mathrm{5}}\mathrm{N}\mathrm{(}\mathrm{a}\mathrm{q}\mathrm{)}\mathrm{\leftrightarrow }{\mathrm{C}}_{\mathrm{5}}{\mathrm{H}}_{\mathrm{5}}\mathrm{N}{\mathrm{H}}^{+}\mathrm{(}\mathrm{a}\mathrm{q}\mathrm{)}+\mathrm{O}{\mathrm{H}}^{\mathrm{-}}\mathrm{(}\mathrm{a}\mathrm{q}\mathrm{)}$

$K_b=\frac{[C_5{H}_{5}N{H}^{+}][O{H}^{-}]}{[C_5{H}_{5}N]}$

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	$K_b=\mathrm{1}.\mathrm{8}\mathrm{\times }\mathrm{1}{\mathrm{0}}^{\mathrm{-}\mathrm{9}}=\frac{x\times x}{.20-x}$
We can assume that x is so small that it will be meaningless by the time we use significant figures.	$\mathrm{1}.\mathrm{8}\mathrm{\times }\mathrm{1}{\mathrm{0}}^{\mathrm{-}\mathrm{9}}\approx \frac{x^2}{.20}$
Solve for x.	$\mathrm{x}\approx \sqrt{.20\times (1.8\times 10^{-9})}=1.9\times 10^{-5}$
Check the assumption that x << .20	$\mathrm{1}.9\times 10^{-5}\ll .20$; so the approximation is valid
Find pOH from pOH = -log [OH-] with [OH-]=x	$\mathrm{p}OH\approx -log(1.9\times 10^{-5})=4.7$
From pH = pKw - pOH,	$\mathrm{p}H\approx 14.00-4.7=9.3$
From the equation for percentage protonated with [HB+] = x and [B]initial = .20,	$\mathrm{p}ercentageprotonated=\frac{1.9\times 10^{-5}}{.20}\times 100\%=.0095\%$

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

Examples

• Alanine,
• Ammonia, NH₃
• Methylamine, CH₃NH₂], C₅H₈O₂

Other weak bases are essentially any bases not on the list of strong bases.

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.
• 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.

See also

• Strong base
• Weak acid

References

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External links

• Explanation of strong and weak bases from ChemGuide
• Guide to Weak Bases from Georgetown course notes
• Article on Acidity of Solutions of Weak Bases from Intute