Question

5. A) If a solution at pH 5 undergoes a 1000-fold increase in [H+], what is the resulting pH?B) If a solution at pH 5 undergoes a 1000-fold increase in [OH-], what is the resulting pH?

Answer 1

REmember that pH is the negatige logarithem of the concentration of protons:

$pH=-\log _{10}\lbrack H^+\rbrack$

when we make the invers of the logarithm:

$\lbrack H^+\rbrack=10^{-pH}$

in this case:

$\lbrack H^+\rbrack=10^{-5}$

if the concentration of protons undergoes a 1000-fold increase the new concentration is:

$\lbrack H^+\rbrack=10^{-5}\times10^3=10^{-2}$

therefore the new pH is calculated as follows:

$pH=-\log _{10}\lbrack H^+\rbrack=-\log (10^{-2})=2$

B.

For the second part we need to remember that [H+] and [OH-] are related according the following equation:

$\lbrack H^+\rbrack\times\lbrack OH^-\rbrack=10^{-14}$

We have calculated before that

$\lbrack H^+\rbrack=10^{-5}$

Then we can calculate [OH-]:

$10^{-5}\times\lbrack OH^-\rbrack=10^{-14}\text{ }\rightarrow\lbrack OH^-\rbrack=10^{-9}\text{ }$

1000 fold that concentration is

$\lbrack OH^-\rbrack=10^{-9}\times10^3=10^{-6}$

Again we use the relation between [H+] and [OH-] with the new value:

$\lbrack H^+\rbrack\times\lbrack OH^-\rbrack=10^{-14}$$\lbrack H^+\rbrack\times10^{-6}=10^{-14}$$\lbrack H^+\rbrack=10^{-8}$

And once again the pH formula:

$pH=-\log _{10}\lbrack H^+\rbrack=-\log (10^{-8})=8$