# Calculating the pH of a Buffer Solution

## The pH of a buffer can be calculated from the equilibrium constant and the initial concentration of the acid.

#### Key Points

• The strength of a weak acid (buffer) is usually represented as an equilibrium constant. The acid-dissociation equilibrium constant, which measures the propensity of an acid to dissociate, is described using the equation: ${ K}_{a }=\frac { { [H }^{ + }][{ A }^{ - }] }{ [HA] }$.

• Using Ka and the equilibrium equation, you can solve for the concentration of [H+]. The concentration of [H+] can then be used to calculate the pH of a solution, as part of the equation pH = -log([H+]).

• The concentration of hydrogen ions can be solved for exactly using the quadratic equation: ${x}^{2}+{K}_{a}x-{K}_{a}[HA]=0$. If Ka >> [HA], the equation simplifies to: $x = \sqrt { { K }_{ a }[HA] }$.

#### Terms

• the state of a reaction in which the rates of the forward and reverse reactions are the same

• to remove one or more protons from a molecule

#### Figures

1. ##### The pH and pOH Scales

The bottom scale is the range of p[H] (red = acid region, blue = basic region). The top scale is the range of p[OH].

2. ##### pH and Buffers Calculations

Hi my name is Jason Matthew from Trinidad and Tobago. This video goes through 3 basic pH and buffer calculations. This video was initially created for my students studying pH and buffers in Level I Biochemistry at the University of the West Indies in Trinidad and Tobago. Please feel free to go through the video and ask as many questions as you like.

3. ##### Sample ICE Table

This ICE Table shows the values for the reaction HA <-> A- + H+

## What Does pH Mean in a Buffer?

In chemistry, pH is a measure of the hydrogen-ion concentration in a solution (Figure 1). The pH of a buffer can be calculated from the concentrations of the various components of the reaction. As mentioned in the previous unit, the balanced equation for a buffer is:

$HA \rightleftharpoons H^+ + A^-$

The strength of a weak acid (buffer) is usually represented as an equilibrium constant. The acid-dissociation equilibrium constant (Ka), which measures the propensity of an acid to dissociate, for the reaction is:

$K_{a} = \frac{\left [H^{+} \right ]\left [A^{-} \right ]}{\left [HA \right ]}$

For more information on equilibrium constants or acid-dissociation constants, see the respective units. The greater the value of Ka, the more the formation of H+ is favored, and the lower the pH of the solution.

## Calculating pH of a Buffer

Figure 2

The pH of a solution of a weak acid depends on the strength of the acid and the other components in the solution. In the simplest case, the weak acid is the only compound in water. In this case, the pH can be found using the concentration of the acid, Ka, to solve for concentration of H+.

Applying the equilibrium to the acid-association expression yields:

${ K }_{ a } = \frac { { [H }^{ + }][{ A }^{ - }] }{ [HA] } = \frac { { x }^{ 2 } }{ [HA]-x }$

Rearranging the above yields:

${ x }^{ 2 }+{ K }_{ a }x-{ K }_{ a }[HA] = 0$

which can be solved for x using the quadratic equation. At equilibrium, x is the concentration of [H+]. Therefore, pH can be solved using the following equation:

$pH = -log(x)$

## A Simplified Method for Calculating Buffer pH

In cases where [HA] is more than 1000 times greater than Ka, the acid will not deprotonate much, and the value of x will be small. Therefore, [HA] - x ≈ [HA]. This simplifies the Ka expression to:

${ K }_{ a } = \frac { { x }^{ 2 } }{ [HA] }$

Solving for x yields:

$x = \sqrt { { K }_{ a }[HA] } = { [H }^{ + }]$

Then the pH can be solved as above. The formulas can be combined to give:

$pH = -log(\sqrt { { K }_{ a }[HA] } )$

but only if [HA] >> Ka.

## Example of Calculating the pH of Buffer Solution

Two buffer solutions are made from an acid HA with a Ka = 1 x 10-5. One solution has a concentration of 0.10 M, and the other has a concentration of 5 x 10-4 M. Calculate the pH for both solutions using both methods described above.

### SOLUTION:

0.1 M Solution

Using the full method gives us the following quadratic:

${ x }^{ 2 }+(1\cdot { 10 }^{ -5 })x-(1\cdot { 10 }^{ -6 }) = 0$

which yields x = 9.95 x 10−3 M and pH = 3.00.

The simplified method gives us the following:

$pH = -log(\sqrt { (1\cdot { 10 }^{ -5 })(0.1) } ) = 3.00$

In this example, [HA] is more than 1000 times greater than Ka, and both methods yield the same result.

0.5 x 10−4 M Solution

Using the full method gives us the following quadratic:

${ x }^{ 2 }+(1\cdot { 10 }^{ -5 })x-(5\cdot { 10 }^{ -9 }) = 0$

which yields x = 6.6 x 10−5 M and pH = 4.18. The simplified method gives:

$pH = -log(\sqrt { (1\cdot { 10 }^{ -5 })(5\cdot { 10 }^{ -4 }) } ) = 4.15$

Here, the results differ by 0.03 pH units. As [HA] becomes closer in value to Ka, then the difference will increase even more.

## ICE Tables: A Useful Tool For Solving Equilibrium Problems

ICE (initial, change, equilibrium) tables are very helpful tools for understanding equilibrium, and as such, for calculating the pH of a buffer solution. Consider, for example, the following problem:

Calculate the pH of a buffer solution that initially consists of 0.0500 moles of NH3 and 0.0350 moles of NH4+, after 30.0 mL of 0.50 M NaOH has been added to the buffer. Note: Ka for the NH4+ is 5.6*10-10.

We know that initially there are 0.0350 moles of NH4+ and 0.0500 moles of NH3. We also know that OH- will react with NH4+ as:

$OH^-+NH_4^+ \rightarrow NH_3 + H_2O$

The decrease in moles of NH4+ will be equal to the increase in NH3, and both will equal the moles of NaOH added. We can therefore determine all three values with one calculation, that of moles of NaOH:

$30.0mL * \frac{1L}{1000mL}*0.5 \frac{mol}{L}=0.015 mol$

Thus, the change in moles of NH4+ is -0.015 mol and the change in moles of NH3 is +0.015 mol. The equilibrium amounts of each can be calculated as the sum of initial and change, leaving 0.0650 moles of NH3 and 0.02 moles of NH4+. Throughout this whole process we have ignored the change in concentration of H+, which is what we need to find pH. This change we will simply define as x, which along with all the initial, change, and final information we can fit into a table.

Sample table: Figure 3

Depending on what is asked by a problem, x can be inserted into any cell in an ICE table, which helps organize data and makes it easier to visualize how to find the unknown variable. In this example, we can find x using the equilibrium concentrations:

$[H_3O^+]=x=5.6*10^{-10}(\frac{.02}{.065})=1.72*10^{-10}$

$pH=-\log(H_3O^+)=9.76$

#### Key Term Glossary

acid
an electron pair acceptor; generally capable of donating hydrogen ions
##### Appears in these related concepts:
acid dissociation constant
quantitative measure of the strength of an acid in solution; typically written as a ratio of the equilibrium concentrations
##### Appears in these related concepts:
balanced equation
The law of conservation of mass dictates the quantity of each element does not change in a chemical reaction. Thus, each side of the chemical equation must represent the same quantity of any particular element. Similarly, the charge is conserved in a chemical reaction. Therefore, the same charge must be present on both sides of the balanced equation.
##### Appears in these related concepts:
buffer
a solution used to stabilize the pH (acidity) of a liquid
##### Appears in these related concepts:
buffers
a weak acid or base used to maintain the acidity (pH) of a solution near a chosen value, prevent a rapid change in pH when acids or bases are added to the solution
##### Appears in these related concepts:
chemistry
The branch of natural science that deals with the composition and constitution of substances and the changes that they undergo as a consequence of alterations in the constitution of their molecules.
##### Appears in these related concepts:
compound
A substance made from any combination elements.
##### Appears in these related concepts:
concentration
the proportion of a substance in a mixture
##### Appears in these related concepts:
constant
Consistently recurring over time; persistent
##### Appears in these related concepts:
deprotonate
to remove one or more protons from a molecule
##### Appears in these related concepts:
dissociation
Referring to the process by which compounds split into smaller constituent molecules, usually in a reversible manner.
##### Appears in these related concepts:
equilibrium
the state of a reaction in which the rates of the forward and reverse reactions are the same
##### Appears in these related concepts:
equilibrium constant
referring to a numerical value derived from the ratio of concentrations of products to reactants of a reversible reaction
##### Appears in these related concepts:
favored
to proceed in a particular direction
##### Appears in these related concepts:
ion
An atom or group of atoms bearing an electrical charge, such as the sodium and chlorine atoms in a salt solution.
##### Appears in these related concepts:
mole
In the International System of Units, the base unit of the amount of substance; the amount of substance of a system that contains as many elementary entities as there are atoms in 0.012 kg of carbon-12. Symbol: mol. The number of atoms in a mole is known as Avogadro’s number.
##### Appears in these related concepts:
pH
the negative of the logarithm to base 10 of the concentration of hydrogen ions, measured in moles per liter; a measure of acidity or alkalinity of a substance, which takes numerical values from 0 (maximum acidity) through 7 (neutral) to 14 (maximum alkalinity)
##### Appears in these related concepts:
a polynomial equation of the second degree.
##### Appears in these related concepts:
solution
A homogeneous mixture, which may be liquid, gas or solid, formed by dissolving one or more substances.
##### Appears in these related concepts:
Solution
A homogeneous mixture, which may be liquid, gas or solid, formed by dissolving one or more substances.
##### Appears in these related concepts:
weak acid
one that dissociates incompletely, releasing only some of its hydrogen atoms into solution