The acid dissociation constant, Ka, of ethanoic acid is 1.78 x 10^-5 at 298K. Given that the concentration of a sample of ethanoic acid is 0.4moldm^-3, calculate its pH at 298K.
Using the acid dissociation equation, Ka = [H+][A-]/[HA]. (where Substitute the known values into the concentration to give 1.78x10-5= [H+][A-]/0.4 . Because the acid is dissociating in solution the acid dissociates in water which is neutral, then [H+] and [A-] must be equal. So we can write: [H+]2/0.4=1.78x10-5. So [H+]2 =7.12x10-6 and [H+] = 2.67x10-3.Substitute [H+] into the pH equation: pH =-log[H+] = -log[2.67x10-3] = 2.57
SR
Related Chemistry GCSE answers
All answers ▸Calculate the mass of copper oxide required to produce 24.95 g of copper sulfate crystals (CuSO4.5H2O).
Iron can be extracted from iron ore by reaction with carbon at very high temperature in a blast furnace. In this process Fe2O3 is reduced to Fe. Explain what is meant by the term reduction.
What are the properties of ionic compounds?
An isotope of Nitrogen has an atomic number of 7 and a mass number of 15. How many protons, neutrons and electrons will each atom have?does