Question

"how many fold difference is the hydrogen ion concentration if the ph is 6.7 is increased to 8.7"

Answer 1

pH of the solution is calculated using the following formula:

`pH=-log{[H^(+)]}`

Here,
`[H^(+)]` is concentration of hydrogen ion.

Initial value of pH is 6.7, calculate concentration of hydrogen ion from this as follows:

`[H^(+)]=10^(-pH)`

Putting the value,

`[H^(+)]=10^(-6.7)=1.9952* 10^(-7)`

Thus, initial concentration of hydrogen ion is
`1.9952* 10^(-7)`.

Now, final pH value is 8.7, calculate concentration of hydrogen ion as follows:

`[H^(+)]=10^(-8.7)=1.9952* 10^(-9)`

Thus, final concentration of hydrogen ion is
`1.9952* 10^(-9)`.

The ratio of final to initial value will be:

`(1.9952* 10^(-9))/(1.9952* 10^(-7))=(1)/(100)`

Thus, if pH increases from 6.7 to 8.7, concentration of hydrogen ion becomes
`(1)/(100)` of the initial value.