What is the pH of pure water at 25.0 °C if the Kw at this temperature is #1.0 x 10-^14#?

1 Answer
Jun 9, 2018

Answer:

#pH = 7.00#.

Explanation:

Note: It's generally pretty well known that the #pH# of pure water at #"25 ºC"# is #7#, but here's what we'd do to get to find that #pH# if we didn't know this:

To find #pH#, we'll need to use two formulas:

  1. #K_w = [H^+]xx[OH^-]#
  2. #pH = -log[H^+]#

First, we need to find the concentration of #H^+# ions in water at this temperature.

We know that pure water is neutral, and that in neutral solutions, the concentration of #H^+# equals the concentration of #OH^-#.

So, effectively:

#K_w = [H^+]xx[OH^-]#
#K_w = [H^+]xx[H^+]# because #[H^+] = [OH^-]# in neutral solution

To find the concentration of #H^+# ions now, we'll just need to square root #K_w#. The question tells us that #K_w = 1.0 xx 10^(-14)#, so:

#1.0 xx 10^(-14) = [H^+]xx[H^+]#

#[H^+] = sqrt(1.0 xx 10^(-14)) = 1.0 xx 10^(-7) M#

Now, we can just use our second equation to calculate the #pH#:

#pH = -log[H^+]#
#pH = -log(1.0 xx 10^(-7) M) = 7.00#

(The number of decimal places in our #pH# value needs to be the same as the number of significant figures in our #[H^+]# value.)