1A) What is the pH of a solution that was 0.75 M acetic acid? Ka for acetic acid is 1.7x10-5. (Note: 0.75 M is the initial concentration of acetic acid.) Set up an ICE table for HC2H3O2(aq)+ H2O (l) C2H3O2-(aq) + H3O+(aq). Remember to cross out any solid or liquid compounds.

1B) Plug the “Equilibrium” (E) terms into the Ka expression:
It appears at first as though you will need to use the quadratic formula to solve for x because the numerator is squared and the denominator is not. However, there is a simple trick that can fix the algebra for these problems. Since Ka is so small, it can be assumed that x is also incredibly small. If x is very small, then 0.75-x will effectively remain 0.75. The “rule of 1000” says that as long as the initial concentration of the acid is at least 1000x larger than the Ka, you can make this assumption. 0.75 is 44,000x larger than 1.7x10-5, so the assumption is valid.

1C) Now that you know the equilibrium concentration of H+, calculate the pH.