# Question 046fa

Oct 3, 2015

$p H = 7$

#### Explanation:

$H C l$ is a strong acid and it will completely dissociate in water. However water will also dissociate to give ${H}^{+}$ and $O {H}^{-}$ ions (auto ionization).

Therefore, from $H C l$ the concentration of ${H}^{+}$ is ${\left[{H}^{+}\right]}_{H C l} = {10}^{- 8} M$ and from ${H}_{2} O$ the concentration of ${H}^{+}$ is ${\left[{H}^{+}\right]}_{{H}_{2} O} = {10}^{- 7} M$.

The pH is calculated from both concentrations ${\left[{H}^{+}\right]}_{T o t a l} = {\left[{H}^{+}\right]}_{{H}_{2} O} + {\left[{H}^{+}\right]}_{H C l} = {10}^{- 7} M + {10}^{- 8} M \approx {10}^{- 7} M$

Therefore the $p H = - \log {\left[{H}^{+}\right]}_{T o t a l} = - \log {10}^{-} 7 = 7$

In this case, the concentration of $H C l$ is too small to change the pH of water. The solution stays neutral.

Dec 27, 2015

Question is wrong and does not make sense.

#### Explanation:

The previous contributor did use the correct method to answer the question and in theory his answer is correct, but the answer makes no sense due to $H C l$ being an acid so must have a pH lower than 7.
Hence, there must be a mistake in the actual question that was asked and this can in fact be proven if one uses the equilibrium constant for water ${K}_{w} = \left[{H}^{+}\right] \left[O {H}^{-}\right] = {10}^{- 14}$.

This would imply that [OH^-]=10^(-14)/(1xx10^(-8)#

$= 1 \times {10}^{- 6} M$

$> \left[{H}^{+}\right]$

This is impossible for an acid it cannot have this.