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In my analytical chemistry course, a question reads:

What is the pH of a 0.10 M solution of THAM? Is it valid to neglect the autoprotolysis of water? Justify and explain your answer by estimating the concentration of $\ce{OH-}$ that comes from just water in this solution.

I'm guessing it has something to do with ionization of water ($\mathrm{K}_\mathrm{w}=1 \times 10^{-14}$), any insight would be greatly appreciated. The calculated pH neglecting autoprotolysis of water is 10.5.

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  • $\begingroup$ For general information - Tris(hydroxymethyl)aminomethane aka TRIS or THAM has a pKa of 8.07 according to Wikipedia. en.wikipedia.org/wiki/Tris $\endgroup$ – MaxW Mar 24 '16 at 23:58
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What auto-ionization means is best explained this way: If you put in 0.10000000 M of NaOH in pure water, how much hydroxide is there at equilibrium? You can imagine that "initially", adding that amount to a neutral solution gives you $\ce{OH^-}$ = 0.10000010, and that $\ce{[H^+]}$ is 0.00000010. However, the ion-product at this point is much greater than $K_w$, so some of the hydronium and hydroxide must react to bring the product down. If you solved the equation $(0.10000010-x)(0.00000010-x) = K_w$, and used $0.10000010-x$ to get $[OH^-]$, you would arrive at the concentration taking into account auto-ionization. You can see that $[OH^-]$ would just be a very minute amount greater than 0.100000000 because x is very small. The x is the result of taking into account the auto-ionization. As you can see, 0.1 dwarfes x, so we could have ignored auto-ionization and called it 0.1. Similarly, the hydronnium concentration would be different than if you had just divided $K_w$ by 0.1 rather than set up an equilibrium equation--but by so very little.

Now for your problem,

Let me first try to find the $K_b$ for THAM. If the pH of a 0.10 M solution is 10.5, then $\ce{[H^+]}$ $3.16*10^{-11}$. Using the mass balance for THAM, (I'll denote THAM as $\ce{B}$ and the conjugate acid as $\ce{BH^+}$),

$\ce{[B]}$ + $\ce{[BH^{+}]}$ = 0.10 M

$\ce{OH^-}$ = $3.16*10^{-4}$ (assuming 298 K) because $\ce{H^+ }$ = $3.16*10^{-11}$

If you neglige the autodissociation of water, in other words, that all of the hydroxide in the solution is due to the THAM, your charge balance equation will be that $\ce{[BH^+]}$ is the same as $[OH^-]$ (by stoichiometry). Then, $\ce{[B]}$ = 0.09968, and $K_b$ is $1.00 * 10^{-6}$.

This should tie everything together:

Before, we had said that $[BH^+] = [OH^-]$. Instead, if you took into account the auto-ionization, you get the equation that $[BH^+] + [H^+ ] = [OH^-]$. Why is it justified to ignore $[H^+]$? Because the solution is so basic, $[H^+]$ is so small that if you tried to take auto-ionization into account, the pH would be roughly the same within quite a few decimal places. The change in $[OH^-]$ that comes about due to auto-ionization is dwarfed by the basicity of the THAM addition, just like the 0.100000 M NaOH example that I used. In other words, if you make the solution intensely basic, you can afford to ignore the effects of auto-ionizaton of water, and similarly for acids. In general, it is OK to ignore the auto-ionization process of water if the solution is more than just slightly acidic/basic. As an interesting note, try to calculate the pH of a $10^{-10}$ M solution of hydrochloric acid without taking into account auto-ionization. You get a pH of 10, which is definitely not right.

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  • $\begingroup$ I upvoted since this is essentially correct. However there are two checks: $$ [\mathrm{B}] >> [\mathrm{B}^+] $$ so that you don't have to solve quadratic equation and $$ [\mathrm{OH}^{-}] \approx [\mathrm{B}^+] $$ so that you can ignore autoionization of water. $\endgroup$ – MaxW Mar 26 '16 at 3:05
  • $\begingroup$ The original question was about the autoionization/protolysis of water. $\endgroup$ – Yunfei Ma Mar 26 '16 at 12:17

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