A person go to an all-you-can eat restaurant but eat a little too much and get heartburn. He returns to his room, looking for a remedy. He takes $3$ teaspoons of milk of magnesia, which contains $\ce{400 mg MgO}$ per teaspoon. What will be the $\ce{pH}$ of the person's stomach after taking the medication, assuming a stomach acid volume of $\ce{100 mL}$ and that the concentration of $\ce{HCl is 0.10 M}$?

I got a pH of 10.15, however, my textbook says the answer is 13.64. Below is my reasoning:

Initially, there are $\ce{30 mmol}$ of $\ce{MgO}$ and $\ce{10 mmol of H+}$, which undergo the following reaction: $\ce{MgO + 2H+ -> Mg^{2+} + H2O}$. After that reaction, $\ce{25 mmol}$ of $\ce{MgO}$ and $\ce{5 mmol}$ of $\ce{Mg^{2+}}$ remain in the solution.

Then, $\ce{MgO}$ reacts with $\ce{H2O}$ to form $\ce{Mg(OH)2}$: $\ce{MgO + H2O -> Mg(OH)2}$. The $\ce{Ksp}$ of $\ce{Mg(OH)2}$ is $1.1*10^{-11}$, so $\ce{[OH-]} = (1.1*10^{-11}/4)^{1/3} = 1.4*10^{-4}$, and the $\ce{pH = 10.15}$. What am I doing incorrectly?

  • 2
    $\begingroup$ 13.64 is way too high. You can't go that alkaline with MgO; if you could, it would be too dangerous to use as a medication in the first place. $\endgroup$ – Ivan Neretin Aug 6 '18 at 5:20
  • $\begingroup$ There are a lot of assumptions which can be made or not made when dealing with this question. Do we assume all activity coefficents to be equal to one or not ? Do we assume that Ka for hydrochloric acid is infinity or not. There are more that we can think about but the best bit is that for neutral water at body temperture the pH is 6 point something. Kw increasees as the temperture goes up, thus the pH of pure water goes down as T goes up. $\endgroup$ – Nuclear Chemist Aug 6 '18 at 10:52
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    $\begingroup$ Assumptions or no assumptions, 13+ is beyond anything reasonable. $\endgroup$ – Ivan Neretin Aug 6 '18 at 16:37
  • $\begingroup$ Not to mention that MgO is quite unreactive and milk of magnesia is $\ce{Mg(OH)2}$, which has a different mass percent of magnesium than $\ce{MgO}$ $\endgroup$ – A.K. Aug 7 '18 at 13:00

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