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06} Balecen phelol abd Ielaoic acja wwhick cic is tncve acidic and explajut wbty. (2+8-[0 pwitls}...

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06} Balecen phelol abd Ielaoic acja wwhick cic is tncve acidic and explajut wbty. (2+8-[0 pwitls}

06} Balecen phelol abd Ielaoic acja wwhick cic is tncve acidic and explajut wbty. (2+8-[0 pwitls}



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Chloroacetic acid, $\mathrm{ClCH}_{2} \mathrm{COOH}$ , has a p $K_{\mathrm{a}}$ of $2.87 .$ What are $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right], \mathrm{pH},\left[\mathrm{ClCH}_{2} \mathrm{COO}^{-}\right],$ and $\left[\mathrm{ClCH}_{2} \mathrm{COOH}\right]$ in 1.25 $M$ $\mathrm{ClCH}_{2} \mathrm{COOH} ?$

Here in this question, solution is basic solution and solution. B is also basic solutions. Solutions. He is a serious solution. And solution is basic solution. Yeah.

Similar to the previous problem. We need to identify what is stronger for part A. We have potassium hydroxide and we have barium hydroxide. They are both strong bases, so they contribute equally by completely dissociating according to their concentrations. Potassium hydroxide is 0.5 Moeller barium hydroxide is 0.15 Moeller, but we get to hydroxide is for every barium hydroxide. So the total hydroxide concentration from both of these strong bases completely dissociating is going to be equal to the contribution from potassium hydroxide at 0.50 Moeller and two times the contribution from barium hydroxide. Because there are two hydroxide for everyone. Barium hydroxide. So two times 20.15 Mueller and we get a total hydroxide concentration of 0.80 Moeller. So the pH is going to be the negative log of the hydro knee, um, concentration, which is going to be the hydroxide concentration we just calculated divided into K W. We take the negative log of that ratio and we get a pH of 12.90 for the next one. We recognize that we have ammonium, which is a weak acid, and we have hydro satanic acid, which is a weak acid. So we have to weak acids, and we want to figure out which weak acid is stronger. So we'll solve for the que a value of ammonium by taking the K B value of ammonia, which we looked up and dividing it into K. W. And we get 5.68 times 10 the negative 10. Then we look up the K a value for hydro satanic acid, and it's essentially the same. 4.9 times tend to the negative 10. So one way to handle this is to assume that both of them are behaving to the same extent. And we can some there to concentrations together to get a total weak acid concentration and then just use an average K value something around five times 10 to the negative 10. Then the hydro knee. Um, concentration would be equal to some average K, a value multiplied by the some concentration of both of the weak acids with essentially equivalent K A values. And we get a hydro knee, um, concentration of about 1.4 times 10 to the negative five, and then we take the negative log of that and get a pH of 4.85 for the next one. We have a solution that contains rubidium hydroxide, which is a strong base, and we have sodium bicarbonate where bicarbonate is AMFA, Terek and can behave is an acid or a base. So it is likely that if we have the strong base rubidium hydroxide in a solution with potassium carbonate that that rubidium hydroxide is going to react with the bicarbonate and produce just carbonate. However, the solution in the solutions manual does not consider that complex reaction. So I'm not going to either. I'm just going to assume that the solution has already reached equilibrium, and these are the concentrations of rubidium hydroxide and sodium carbonate after reaction has occurred. If that's the case, then we focus on the stronger of the two, which is rubidium hydroxide, and we get a hydroxide concentration equal to the rubidium hydroxide concentration of 0.75 So the pH will be equal to the negative log of the hydro knee, um, concentration, which is the hydroxide concentration coming from rubidium hydroxide divided into K. W. And we get 12.88 then for D. We see that we have a solution of the strong acid perk Lorik acid and a solution of the strong base potassium hydroxide. Now, in this case, in order to get the correct answer, we do need to recognize that there is a reaction between these two and that some of the potassium hydroxide is going to consume well, all of the potassium hydroxide is going to consume some of the perk Lorik acid. So the concentration than of HCL 04 is going to be the 40.88 minus what reacts with potassium hydroxide. 0.0 to 2 and we get a concentration then of the strong acid, um, folkloric asset of 0.66 So the ph, then it's going to be the negative log of 0.66 or 1.18 Then for the last one, we see that we have a solution of sodium. Ah, hypochlorite hypochlorite came from the acid Heiple Cloris acid. So itself is a base, and then potassium iodide does not do anything to affect the pH of the solution. Potassium coming from the strong based potassium hydroxide and iodide coming from the strong acid hydro hydro I otik acid. So we focus on the c l o minus to determine the pH of the solution. We can calculate the KB of Cielo minus by dividing DK a value of HIPAA Cloris acid into K W to get a K value of 3.45 times 10 the negative seven. Well, then get our hydroxide concentration by taking the K B value multiplied by the concentration of Hippo. Cloris, I'm sorry. Multiplied by the concentration of hypochlorite ceelo minus at 0.115 Take the square root of that and we get 1.99 times 10 to negative four. This is definitely less than 5% of the 50.115 So we don't need the quadratic we can then get the pH by taking the negative log of the hydro knee. Um, concentration, which is the hydroxide concentration we just determined divided into K W. And we get 10.30

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.

As we know that, as we know that I do flurry cosine, I do. Hello, Dick acid. Age is stronger acid. It is stronger acid, then boric acid, then boric acid. Therefore, according to the given option in this problem ups and be of 78 got it and said. But this problem I do glory Cassidy's stronger acid than boric acid.


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