This problem uses the
relationship between Ka and the concentrations of the ions. Calculations are as follows:
1.9 x 10-5= x^2 / (0.25 - x)
x is very small and the denominator is approximately equal to 0.25. Thus, x is 2.2 x 10^-3
pH = -log (2.2 x 10^-3) = 2.66
