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Why is the pOH of 1 molar HCl 14 in an aqueous solution?

I was learning about strong bases and strong acids in chemistry within this context:

We have 1L of pure water at 25°C, and the pH of water is 7 and the pOH of water is 7, so it is a neutral solution.

Then we add 1 mole of HCl into the water. HCl, because it is such a strong acid, disassociates completely in the water. After some basic calculations, we found that the pH of HCl was 0 and the pOH was 14, but the way he showed this was using the water's pH and pOH - does this make sense?

So we found that the pH of water is 0, and then he did something funny and got a pOH of 14. Why is the pOH of this solution 14?

Here is the video, video mark, 10:02: Khan Academy - pH & pOH of Strong Acids and Bases

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    $\begingroup$ pH + pOH = 14. This follows from $k_w = 10^{-14}$. $\endgroup$ – aventurin Jun 16 '16 at 15:15
  • $\begingroup$ @aventurin keep in mind this is at STP; at higher temperatures $pK_w$ will be lower. $\endgroup$ – Neil Chowdhury Jun 13 '18 at 23:49
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In the moment from the video that you referred to, Sal was using the equation $\rm pK_w=pH+pOH$ to find the $\rm pOH$ of the solution. $\rm pK_w$ is equal to 14, since $K_w$ is equal to $1\times 10^{-14}$, giving us the new equation $14=\rm pH+pOH$. We know that the $\rm pH$ of a 1M solution of $\ce{HCl}$ is zero from earlier on in the video. This gives us that $14=0+\rm pOH$, or in words, that the $\rm pOH$ must be equal to 14.

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In case the $\mathrm{p}$'s are confusing, why don't we take a step back and think about what $K_\mathrm{w}$ means?

The reaction of water's autoionisation is $\ce{H2O(l)->H+(aq) + OH-(aq)}.$ This gives the following expression for $K$, the equilibrium constant:

$$K_\mathrm{w}=[\ce{H+}][\ce{OH-}]$$

Remember that $K_\mathrm{w}$ will be a constant at standard conditions (actually, it is $1.0\times10^{-14}$ at STP). But let's try to write this equation in terms of $\mathrm{p}K_\mathrm{w}$, remembering our log rules!

$$-\log K_\mathrm{w}=-\log ([\ce{H+}][\ce{OH-}])$$ $$\mathrm{p}K_\mathrm{w}=-\log [\ce{H+}] -\log[\ce{OH-}]$$ $$\mathrm{p}K_\mathrm{w}=\ce{pH}+\ce{pOH}=14$$

Now you should understand where this magical equation comes from. Since water is approximately neutral, it will have $\ce{pH}=\ce{pOH}=7$ at standard temperatures.

You got that the $\ce{pH}$ of $\ce{HCl}$ was $0$ from taking the negative log of $\pu{1M}$. That means that $\ce{pOH}=pK_w-\ce{pH}=14-0=\boxed{14}$.

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