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10volume5 Eqles -5662Serizs]Linear (Series2|Linear {5eries31Find pH at half equlvalence pt,...

Question

10volume5 Eqles -5662Serizs]Linear (Series2|Linear {5eries31Find pH at half equlvalence pt,

10 volume 5 Eqles - 5662 Serizs] Linear (Series2| Linear {5eries31 Find pH at half equlvalence pt,



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Determine the pH of each solution.
a. $1.0 \times 10^{-2} \mathrm{M} \mathrm{NaOH}$
b. $1.0 \times 10^{-3} \mathrm{M} \mathrm{M} \mathrm{KOH}$
c. $1.0 \times 10^{-4} \mathrm{M} \mathrm{LiOH}$

Chapter 15 problem. Nine. Ask us to determine the pH of various solutions. So here. Remember that pH is equal to the negative log of your concentration of hydrogen atoms. And also remember. But if your concentration is given in the terms up one times 10 to the negative value, that value will actually be your pH. If this is your 100 I concentration, um, and if you actually plug in this value into here, you'll see that that's that's correct. It's essentially the same thing. This is just a shortcut method. Or if your concentration is given in this form, the exponents negative of it will be your pH. So that's the concept that we will use here. So our first part here, part a tells us that we have 1.0 times 10 to the negative three Moeller of H C O. Now, one thing to keep in mind here is this equation works as long as we only have one hydrogen in each molecule, our asset so that applies here. So the ph of this one would be three. Next, let's move on to be and were given the same concentration here tend to the negative three Moeller of h and 03 Again, we only have one hydrogen per molecule So that's also a peach of three. Next, if you look at sea, were given the same kind of format here. So 1.0 times 10 to the negative five Moeller of h I. Once again, just check. Make sure you only have one hydrogen per molecule. If you do, your pH is going to be the negative exponents here and our last one. We have 1.0 times 10 to the negative for Mueller of h br one hydrogen per molecule. Great. So you can just go ahead and use the excrement.

Guys, we're going to go from 1 12 Chapter 14 s. So they want to calculate the pH of 5.0 time sent to *** third Moeller of h u S 04 we're gonna do. So it's us. 04 will yield H plus plus H s O for minors. And since this and we don't to use cave A values because issue it's a force a strong acid, we don't have toe. We don't have to look at the second reaction of issue Uh, h s 04 h plus an s +04 because that's a very weakly acidic, acidic reaction. So it's not gonna have much Bear is not really gonna affect our It's not really gonna affect our overall answer that we're gonna use negative Bess. They get a vlog of h plus concentration is gonna equal r P h. That's going to equal negative log of 5.0 times 10 to the negative. Third Moeller been equal. Ph. O. P. H is going to equal oh, two 0.3

To determine the ph of a solution that contains assault. We need to identify whether the cat ion or the an ion is an acid or a base. Cat ions are typically acids and an ions are typically based bases, but not all of them. And ions that come from strong acids do not affect the pH of at all and are not significant bases. And cat ions that come from a strong base does not affect the ph and is not a significant acid. So for the first one we have ammonium chloride, chloride came from the strong acid hcl. So chloride doesn't affect the pH but NH four plus came from the weak base ammonia and therefore ammonium. NH four plus is a significant weak acid. So to solve for the ph of this solution, we need to know the K. A. Of ammonium. To get the idea of ammonium, we can look up the KB of ammonia. NH three K A will then be equal to the KB of ammonia divided into K. W. Recognizing that K. A. Multiplied by KB. Gives us kW. So K. A. Will be equal to KW divided by K. B. And we get A. K. A. Value for ammonium of 5.68 times 10 to the negative 10. The K. A. For any weak acid can be used to determine ph by setting it equal to the hydro knee. Um concentration squared divided by the concentration of the weak acid minus the hydro knee um concentration. Many people will assume that the hydro knee um concentration is small with respect to the original concentration of the weak acid. And drop it from the calculation. Then the hydro knee um concentration can simply be calculated by taking the square root of this concentration multiplied by ca or you can choose to use the quadratic which I did and salt for the hydro knee um concentration as 7.54 times 10 to the negative six. This is the hydro knee um concentration not the hydroxide. So this is a church three plus. Now to solve for ph we simply take the negative log of the hydro knee um concentration that we determined here. And ph for a 0.1 Mueller ammonium chloride solution is 5.12 For the next one we have the salt, sodium. Acetate, sodium came from the strong based sodium hydroxide. So it does nothing to affect the ph acetate did not come from a strong acid. It came from the weak acid acetic acid. Therefore acetate is a significant weak base. To solve for the ph of this solution, we need to know the KB of acetate. To get the KB of acetate, we need to divide the K. A. Of acetic acid into KW. So the KB of acetate will be equal to the K. Of acetic acid 1.8 times 10 to the negative five divided into K. W. 1.8 times 10. The negative 14. And we get a KB of 5.56 times 10 to the negative 10. Knowing the KB. And for any weak acid, knowing KB can be set equal to the hydro hydro knee. Um Sorry the hydroxide concentration squared divided by the concentration of the week base. In this case acetate at 0.1 Mueller minus the hydroxide concentration. Again many people choose to drop the hydroxide from the denominator, assuming that it is small with respect to 0.1. And then calculate the hydroxide concentration by multiplying 0.1 by KB and taking the square root. Or you can use the quadratic formula as I did. And the hydroxide concentration is 7.45 times 10 to the negative six moller. So to get the ph we need to take the negative log of the hydro knee um concentration, not the hydroxide concentration. So to get the hydro knee um concentration from the hydroxide concentration, we divide the hydroxide concentration into KW. This gives us our hydro knee um concentration that we then take the negative log of to get a ph of 887 The last salt is sodium chloride, sodium came from the strong base sodium hydroxide. So it will not affect the ph chloride came from the strong acid hydrochloric acid. So it also does not affect the ph So in theory, without any other discussion, the sodium chloride has a ph equal to that of water which is seven. This is not entirely true for things that we have not yet discussed, such as temperature changes, which can affect ph and ionic strength of a solution, which can affect PH. But for the purpose of this textbook, if it is assault that came from a strong acid and strong base than PH is seven.


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