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Title explains all. I have been stuck on this for an hour and for some reason cannot understand it. I have tried to do an ICE table but get stuck halfway as I do not know whether I would use the w/v to calculate the molarity and use that to neutralize the acid with the base and find the "free" H+?
I appreciate any insight!
In general the total equilibrium concentration of any acid HA is
$$c_a=\mathrm{[HA]_e+[H^+]_e - [OH^-]_e}\tag{1a}$$
and for a base
$$c_b=\mathrm{[B]_e -[H^+]_e + [OH^-]_e\tag{1b}}$$
When there is a strong acid and base the pH is not always what is expected due to the equilibrium $\mathrm{H_2O \leftrightharpoons H^+ + OH^-}$. This is particularly the case around pH = 7. To calculate the pH of a strong base start with eqn 1(a) but for the strong base the equilibrium amount of base $\mathrm{[B]_e}=0$ and so the equation to solve is
$$c_b=\mathrm{ -[H^+]} + K_w/\mathrm{[H^+]}$$
where $K_w=\mathrm{[H^+][OH^-]} = 10^{-14}$. This equation is, letting $x =\mathrm{[H^+]}$, $x^2+c_bx-K_w=0$ and solving for a concentration of, for example, $5\cdot 10^{-8}$ M produces $\mathrm{[H^+]}=7.8\cdot 10^{-8}$ or pH = 7.11.
In the case of a strong acid [HA] = 0 and equation (1a) solved but once the $\mathrm{[OH^-]}$ is known $K_w$ can be used to find $\mathrm{[H^+]}$
The figure shows the pH, for example, of a KOH solution at a range of concentrations. Around 10-7 molar the pH is greater than that anticipated. In a strong acid the pH is smaller than anticipated (more acidic) around this concentration.
Figure. The pH of a strong acidic and strong basic solution (e.g. KOH ) at a range of concentrations.
The conjugate base of a strong acid and the conjugate acid of a strong base are both neutral, and do not contribute to the pH of the mixture.
However, the $\ce{H+}$ ions in the strong acid and the $\ce{OH-}$ ions in the strong base react completely to form water when the strong acid and strong base are mixed: $\ce{H+ + OH- -> H2O}$. To perform the stoichiometry calculation, the number of moles of $\ce{H+}$ ions and the number of moles of $\ce{OH-}$ ions in the acid and base solutions, respectively, can be determined by multiplying the $wt/v\%$ by the volume of the respective individual solution, and dividing the wt by the molar mass of the acid or base.
Once the reaction goes to completion, the concentration of any excess unreacted $\ce{H+}$ or $\ce{OH-}$ can be used to calculate the resulting $\ce{pH}$.
