How the pH changes in water with a stong acid or base (but mainly acid)
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How the pH changes in water with a stong acid or base (but mainly acid)
In one of last week's lectures Dr. Lavelle mentions that adding a weak acid and calculating the concentration of H30+, the pH of a solution in water does not actually change that much. I was wondering about a similar concept reagrding strong acids and abses. Why is the concentration of H+, for example, in calculating the pH of a solution with a newly added strong acid, not just the concentration before plus the concentration of the acid (since its strong it dissociates entirely)? This is assuming monoprotic
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Re: How the pH changes in water with a stong acid or base (but mainly acid)
Hi!
I feel as if I'm not quite getting your question, as to me, when you have to calculate a strong acid's dissociation in water, it's just the molarity of said strong acid (because as you said, you're assuming monoprotic and also that strong acids dissolve completely). I'm either misreading, or your concept is totally fine and there's nothing to be confused over.
edit: just read the part about adding the hydronium concentration of the strong acid to the original hydronium concentration. i was under the impression that calculating the pH of a strong acid (monoprotic) dissolved in water is as simple as seeing what the acid's molarity is and converting that to the hydronium ion concentration? not too sure what you mean by adding it to the prior concentration.
I feel as if I'm not quite getting your question, as to me, when you have to calculate a strong acid's dissociation in water, it's just the molarity of said strong acid (because as you said, you're assuming monoprotic and also that strong acids dissolve completely). I'm either misreading, or your concept is totally fine and there's nothing to be confused over.
edit: just read the part about adding the hydronium concentration of the strong acid to the original hydronium concentration. i was under the impression that calculating the pH of a strong acid (monoprotic) dissolved in water is as simple as seeing what the acid's molarity is and converting that to the hydronium ion concentration? not too sure what you mean by adding it to the prior concentration.
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Re: How the pH changes in water with a stong acid or base (but mainly acid)
By "concentration before" are you referring to the [H3O+] due to the autoprotolysis of water? If that's the case, technically, you are correct. For strong acid HA, the equilibrium concentration of H3O+ would be the sum of [HA] and the [H3O+] before you added it. However, we know water at 25°C has [H3O+] = 10^-7 M. Say you have [HA]=0.02 M. Adding 10^-7 to 0.02 makes no appreciable difference--you just get 0.02. So we just disregard the [H3O+] before.
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Re: How the pH changes in water with a stong acid or base (but mainly acid)
It usually only makes a difference when the [H3O+] increases by a small amount (around 10^7 or less). However, with the values we are dealing with, in strong acids and bases this will usually not be the case. Usually [H3O+] will be much higher than the [H3O+] already present in water, making it null and void. But in the case of weak bases and acids that only increase the [H3O+] by a very small mount (10^-10, 10^-8 etc.) the hydronium ions already present in water completely outweigh any change the weak bases or acids will change it/
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