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)*10^3=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*\lbrack OH^-\rbrack=10^(-14)`

We have calculated before that

`\lbrack H^+\rbrack=10^(-5)`

Then we can calculate [OH-]:

`10^(-5)*\lbrack OH^-\rbrack=10^(-14)\text{ }\rightarrow\lbrack OH^-\rbrack=10^(-9)\text{ }`

1000 fold that concentration is

`\lbrack OH^-\rbrack=10^(-9)*10^3=10^(-6)`

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

`\lbrack H^+\rbrack*\lbrack OH^-\rbrack=10^(-14)`
`\lbrack H^+\rbrack*10^(-6)=10^(-14)`
`\lbrack H^+\rbrack=10^(-8)`

And once again the pH formula:

`pH=-\log _(10)\lbrack H^+\rbrack=-\log (10^(-8))=8`