In general, how much does the pH of a NaOH solution change when diluting with water ? If the dilution process was continued indefinitely, what would the pH of t

Question

In general, how much does the pH of a NaOH solution change when diluting with water ? If the dilution process was continued indefinitely, what would the pH of the NaOH solution eventually reach ?

Answer 1

The answer to the first question will depend upon the extent of dilution of each step. However, you can express this relationship as a function of the volume of H2O added (V'):

 

[NaOH]f = [NaOH]i(Vi)/(Vi+V')

 

where [NaOH]f and [NaOH]i represent the final and initial concentrations of NaOH and Vi represents the initial volume.

 

Since the pH = 14 - pOH and pOH = -log[OH-], then

 

pH = 14 + log{[NaOH]i(Vi)/(Vi+V')}

 

Since Vi + V' > Vi, the concentration of NaOH will decrease and eventually will be less than 1. At this point the logarithm will be negative and will start becoming less and less basic, as you would expect.

 

There is one little detail that we haven't yet mention and that is that the total hydroxide concentration is a combination of the concentration of NaOH and that produced by water hydrolysis ([OH-]w) or:

 

[OH-]f = [NaOH] + [OH-]w

 

At [NaOH] concentrations greater than micromolar, [OH-]w is insignificant, but the more you dilute NaOH you eventually reach a point where [OH-]w is not only significant, but the dominant concentration. In other words, under infinite dilution the [OH-] becomes 1x10^-7 M and the pH = 7.