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How to add pH's |
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| Aug1-12, 06:19 PM | #1 |
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How to add pH's
A flask contains 500 mL of an acid with a pH of 4. If 500 mL of water is added, what will the pH of the new solution be?
I don't want to know the answer please* I'm just wondering if it is as easy as adding the two pH's ie. 7 + 4 or do I have to work out the H+ of each ie. 10-7 and 10-4 and then add, multiply or divide those. And why? |
| Aug1-12, 07:55 PM | #2 |
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pH is defined by the following equation:
[itex]\large pH = -log[H^{+}][/itex] It's not quite as simple as adding the pHs, but when you halve the concentration of H+ (by doubling the volume of the solution), the new pH (which we'll call pH2) is: [itex]\large pH_{2} = -log\frac{[H^{+}]}{2}[/itex] Which, by laws of logarithms, is equivalent to: [itex]\large pH_{2} = -log([H^{+}])-log(2)[/itex] Since log(2)=-0.301 (approximately), we can say: [itex]\large pH_{2} ≈ pH_{1}+0.3[/itex] where pH1 is the pH before the water was added. |
| Aug1-12, 07:58 PM | #3 |
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pH is not a linear function you can go on adding, it is a logarithmic function. Its addition depends totally upon addition of [H+]
First, we have to calculate the initial concentration of H+. It is 10-4. Now, we diluted it with equal amount of water, so the concentration becomes half. Note that water also consist [H+] = 10-7, but it can be neglected in front of [H+] of initial solution. Now pH = -log10[H+]; [H+] = 5 x 10-5 pH comes out to be 4.3010 |
| Aug2-12, 11:46 AM | #4 |
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How to add pH's |
| Aug2-12, 12:47 PM | #5 |
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No No No... Diluting Acidic solution will increase the pH. See, pH is MINUS Log of concentration of H+. If we dilute the solution, the concentration decreases, its log decreases, but pH increases. (see negative sign).
Here we halved its concentration... So the pH increases. By calculating the -log of 10-4/2; we get 4.3010, not 3.699. 3.699 is the pH of the solution if you actually concentrate it. (eg. like boil the solution until half the initial volume remains). Remember; Dilution increases pH in acidic solutions, decreases pH in basic solution. |
| Aug2-12, 04:45 PM | #6 |
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I should mention that the equations I made will only work for an acidic solution. With a neutral or basic solution, the addition of water does not halve the H+ concentration as we assumed in the equations. The assumption we made is accurate to 2 significant figures for any pH below 6, and it's accurate to 4 sig figs below pH 5. |
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