Question 415314
Inserting {{{10^(-4)}}} into the pH formula we get:
{{{-log((10^(-4)))}}}
Now we can use a property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, to move the exponent of the argument out in front:
{{{-(-4)log((10))}}}
Since log(10) = 1 by definition this becomes:
-(-4)(1)
which simplifies to:
4
The pH of the tomato juice is 4.