pH calculations with examples

Strong acids are known for their tendancy to fully dissociate. This that in order to calculate the pH of a strong acid solution they must use the equation :  

 [H+]= [HA]

After using this equation to calulcate the hydrogen ion concentration use the equation pH= -log(10)[H+] 

Example: what is the pH of a solution with a HCL concetration of 2.45x10-3 moldm-3?

1. HCl--------> H+   + Cl-

2.pH= -log(10)[2.45x10-3] = 2.61

Weak acid only partically dissociate and as such a different set of equations are need to be used to calculate the pH.

When calculating the hydrogen ion concentration the equation that is needed is:

Ka=  [H+]^2 / [HA], This equation needs to be rearranged to Calculate [H+].

  [H+]=  square root (HA x Ka).

Example-  calculate the pH of 0.0245 mol dm-3 ethanoic acid, CH3COOH, at 25 degree celcius, where Ka= 1.7x10-5 mol dm-3.

1. [H+]=  square root ( 0.0245 x 1.7x10-5) = 6.45x10-4

2. pH= -log(10)[6.45x10-4] = 3.19 

Buffer solutions act to minimise any changes in pH when a small amount of acid or base is added.

The equation used to calculate the Ka ( acid dissociate constant) is, 

Ka= [H+] x [A-] / [HA] 

This can be rearranged to form the equation  [H+]= Ka x [HA]/ [A-].

Example-  calculate the pH of a buffer solution contains 0.1 moldm-3  CH3COOH and 0.300 moldm-3 of CH3COONa. Ka= 1.74x10-5.

1. [H+]= 1.74 x10-5 x 0.1/ 0.3 = 5.8 x10-6

2. pH= -log(10)[5.8x10-6] = 5.24  

To calculate the pH of a strong base the equation involving the ionic product of water is used. The ionic product of water is a constant- 1x 10-14.

Kw= [H+] x [OH-]  this equation needs to be rearranged to make [H+] the subject.  

[H+] = Kw/ OH-

example-  what is the pH of sodium hydroxide with a concentration of 0.02 mol dm-3, using the ionic product of water.

1. [H+] = 1x10-14 / 0.02= 5x10-13 

pH= -log(10) [5x10-13]= 12

The Henderson- Hasselbach equation is a very useful equation which can be a quicker method of calculating the pH of a buffer solution.

pH= pKa + log (A-/ HA)

Example-  Calculate the pH of a buffer solution with a pKa of 1.75 x 10-5, 0.03 moldm-3 of CH3COONa and 0.02 moldm-3 of CH3COOH.

1. pH= 1.75x10-5 + log ( 0.03/0.020)= 0.176