Question

Is it possible for acids to have a pH above 7? If so, under what circumstances would that be?

Answer 1

Hint: The pH scale is used to determine the strength of the acidic or alkaline solutions. Acidic solutions mostly have a lower pH than the basic or alkaline solutions. More the amount of ${[H]^ + }$ ion the lower the value on the pH scale.

Complete answer:
The acidity of a solution depends on the concentration of the ${[H]^ + }$ ion. The pH is the most convenient method to measure this as it has a wide range of numbers involved.
The formula for finding the pH is: $pH = - \log [{H^ + }]$
It can also be written as: $[{H^ + }] = {10^{ - pH}}$
It is typically said that the pH scale ranges from 0 to 14. A solution with the pH of 7 is said to be neutral. If the pH is less than 7 it is acidic and if it is more than 7 then it is basic/alkaline. This condition is satisfied at standard laboratory conditions. In other conditions it works differently. Let us first understand the ionisation of water:
${H_2}{O_{(l)}} \rightleftharpoons H_{(aq)}^ + + OH_{(aq)}^ - $
The equilibrium constant for water ${K_w}$ can be given as: ${K_w} = [H_{(aq)}^ + ][OH_{(aq)}^ - ] = 1.00 \times {10^{ - 14}}mo{l^2}{l^{ - 2}}$at standard temperature of ${25^ \circ }C$. Note that in pure water the concentration of ${[H]^ + }$ is same as that of $[O{H^ - }]$ . Now, the value of ${K_w}$ will become; ${K_w} = {[H_{(aq)}^ + ]^2} = 1.00 \times {10^{ - 14}}$
$[H_{(aq)}^ + ] = \sqrt {1.00 \times {{10}^{ - 14}}} = 1.00 \times {10^{ - 7}}mol/L$
Substituting in the pH formula we get the pH as; $pH = - \log (1.00 \times {10^{ - 7}}) = 7$
This gives us the neutral pH at standard temperature. Here’s where the problem starts, the neutral point is at ${25^ \circ }C$ . We know that equilibrium constant is temperature dependant. To prove this let us consider the endothermic process of auto-ionisation of water.
${H_2}{O_{(l)}} \rightleftharpoons H_{(aq)}^ + + OH_{(aq)}^ - $
If we raise the temperature, according to the Le Chatelier’s principle the equilibrium would shift towards the right. This would increase the dissociation hence increasing the value of ${K_w}$ . Experimentally the value of ${K_w}$ at a higher temperature of ${40^ \circ }C$ would be ${K_w} = 2.916 \times {10^{ - 14}}mo{l^2}{L^{ - 2}}$
Therefore, ${K_w} = {[H_{(aq)}^ + ]^2} = 2.916 \times {10^{ - 14}}$
$[H_{(aq)}^ + ] = \sqrt {2.916 \times {{10}^{ - 14}}} = 1.707 \times {10^{ - 7}}mol/L$
$pH = - \log (1.707 \times {10^{ - 7}}) = 6.77$
Thus the point of neutrality has been dropped from 7 to 6.77. The pH falls as the temperature rises, this doesn’t mean that water has become more acidic. All throughout the water was neutral itself. The condition for neutrality should be $[H_{(aq)}^ + ] = [OH_{(aq)}^ - ]$
Similarly, for acidity the condition should be $[H_{(aq)}^ + ] > [OH_{(aq)}^ - ]$ and for alkalinity $[H_{(aq)}^ + ] > [OH_{(aq)}^ - ]$
Concluding that, any acidic solution can have a pH greater than 7 and still be acidic, similarly any alkaline solution can have pH less than 7 and still be basic.

Note:
The standard pH is very helpful in determining the pH at room temperature. The 0 to 14 scale is still applicable but only at room temperature. Remember that Strong acids react with weak bases from acidic salts as they have high ${[H]^ + }$ concentration. Similarly for strong bases and weak acids.