A solution is created by mixing $250\,\mathrm{mL}$ of $1\,\mathrm{M}$ $\ce{HCl}$, $250\,\mathrm{mL}$ of $1\,\mathrm{M}$ $\ce{CH3COOH}$ and $500\,\mathrm{mL}$ of $1.5\,\mathrm{M}$ $\ce{CsOH}$.

The $K_\mathrm{a}(\ce{CH3COOH}) = 1.8 \times 10^{-5}$

What is the $\mathrm{pH}$ of this solution?

I am stuck between $\ce{HCl}$ and $\ce{CsOH}$ since they are both a strong acid and base (respectively) meaning that they should both have a substantial affect on the $\mathrm{pH}$.

Also when determining what the $\mathrm{pH}$ is, which species would we use in our ICE table?


2 Answers 2


CsOH is a strong base.

0.5 L X 1.5 M = 0.75 moles of hydroxide.

Similarly, there are 0.5 moles of monoprotic acid.

After mixing there will be 0.25 moles of hydroxide in 1 liter of solution.

In other words [OH-] = 0.25M

pH is approximately 14 + log (0.25) = 13.4


You need to provide the concentration of acetic acid. However, if I were to guess, I would say that the hydrochloric acid completely reacts with the cesium hydroxide, and the remaining cesium hydroxide reacts with the acetic acid. Finding the quantity of acetic acid, which probably was more than the concentration of cesium hydroxide, you could use hendersen hassebach.


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