Calculation of the pH of a buffer solution
Calculate the pH of a buffer solution formed by adding 20.00 cm3 of 0.10 moldm-3 NaOH to 40.00 cm3 of the weak acid, HX, which has a concentration of 0.20 moldm-3 and a Ka value of 5 x 10-6.
Get the equation right: HX + NaOH ? NaX + H2OK
Quote the Ka expression:
Ka = [H+][X-] / [HA]
Calculate the initial moles of acid:
Moles of acid at start = 40.00 x 0.2 / 1000 = 0.008 moles
Calculate the initial moles of alkali:
Moles of alkali at start = 20.00 x 0.1 / 1000 = 0.002 moles
How much acid (HX) is left over?
Moles of acid left over = moles of acid at the start – moles of alkali added
Moles of acid left over = 0.008 – 0.002 = 0.006
What is final concentration of the acid (HX)?
Concentration of acid = moles of acid / volume in dm3
(don’t worry about V unless you have to because it will probably cancel out)
[HX] = 0.006 / V
How many moles of X-? Moles of X- = number of moles of alkali used up.
Moles of X- = 0.002
What is final concentration of the X-?
Concentration of X- = moles of alkali used / volume in dm3
(don’t worry about V unless you have to because it will probably cancel out)
[X-] = 0.002 / V
Find [H+] by rearrangement and put in the numbers:
[H+] = Ka [HA] / [X-] [H+] = 5 x 10-6 x (0.006 / V) / (0.002 / V)
V cancels and so….. [H+] = 5 x 10-6 x 0.006 / 0.002 [H+] = 1.5 x 10-5
pH = -log10[H+]
pH = -log10 (1.5 x 10-5)
pH = 4.82 (always give pH to 2 dp)
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