In Example 16-9, rather than use the quadratic formula to solve the quadratic equation, we could have proceeded in the following way. Substitute the value yielded by our failed assumption—x = 0.0010—into the denominator of the quadratic equation; that is, use (0.00250 – 0.0010) as the value of [CH₃NH₂] and s1olve for a new value of x. Use this second value of x to re-evaluate [CH₃NH₂]: [CH₃NH₂] = (0.00250 - second value of x). Solve the simple quadratic equation for a third value of x, and so on. After three or four trials, you will find that the value of x no longer changes. This is the answer you are seeking. 

(a) Complete the calculation of the pH of 0.00250 M CH₃NH₂ by this method, and show that the result is the same as that obtained by using the quadratic formula. 

(b) Use this method to determine the pH of 0.500 M HClO₂.

Example 16-9

What is the pH of a solution that is 0.00250 M CH₃NH₂(aq)? For methylamine, K_b = 4.2 x 10^-4.