Calculate the pH of a 0.50 M HF solution at 25°C. The Ka is 7.1 x 10-4, and the ionization of HF is given by HF (aq) H+(aq) + F-(aq).

The students use a table to organize the data to solve this problem.

Then use the equilibrium constant to find x.

One could use the quadratic formula to solve this equation, but an approximation could be made to solve the problem more easily. Since HF is a weak acid we could reason that x must be small compared to 0.50. Thus we make the approximation 0.50 – x ≈ 0.50.

Now can be written as .

Solving for x we find x = 0.109M. Before calculating the pH of the solution, it is wise to determine if the approximation was legitimate. If x is less than 5% of the number it was subtracted from, then the approximation is valid.

Thus the approximation is valid. The pH can be calculated as follows.