Question

The pH of solution is 6.0 . To this solution, sufficient acid is added to decrease the pH to 2.0. The increase in hydrogen ion concentration is:

• A. ${10}^{0.5}$ times.
• B. ${10}^6$ times.
• C. ${10}^4$ times.
• D. ${10}^2$ times.

Answer 1

The correct option is C ${10}^4$ times.

Case-1: $pH=6\Rightarrow -lo{g}_{10}\left[H^{+}\right]=6\left[H^{+}\right]={10}^{-6}.....\left(1\right)$
Case-2: $-lo{g}_{10}\left[H^{+}\right]=2$
$\left[H^{+}\right]={10}^{-2}.....\left(2\right)$
By dividing equation (2) by (1) we get, $[H^{+}{]}_{increase}$ as,
$\frac{{10}^{-2}}{{10}^{-6}}={10}^4=10000times$
Hence, the increase in Hydrogen ion concentration is 10000 times.


Theory:
Hydronium ion concentration in molarity is more conveniently expressed on a logarithmic scale introduced by Sorenson known as pH scale.
pH of a solution is the negative logarithm to the base 10 of the activity ($a_{H^{+}}$) of $H^{+}$ ions.
pH and pOH :
$pH=-lo{g}_{10}\left[H^{+}\right]=lo{g}_{10}\frac{1}{\left[H^{+}\right]}$,$pOH=-lo{g}_{10}\left[O{H}^{-}\right]=lo{g}_{10}\frac{1}{\left[O{H}^{-}\right]}$
In general for any quantity $X$ we can define pX as $pX=-lo{g}_{10}X$