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,

$[H^{+}]=10^{-6.7}=1.9952\times 10^{-7}$

Thus, initial concentration of hydrogen ion is $1.9952\times 10^{-7}$.

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

$[H^{+}]=10^{-8.7}=1.9952\times 10^{-9}$

Thus, final concentration of hydrogen ion is $1.9952\times 10^{-9}$.

The ratio of final to initial value will be:

$\frac{1.9952\times 10^{-9}}{1.9952\times 10^{-7}}=\frac{1}{100}$

Thus, if pH increases from 6.7 to 8.7, concentration of hydrogen ion becomes $\frac{1}{100}$ of the initial value.