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Calculate the change in pH when 3.00 mL of 0.100 M HCI(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH;C (aq) and 0.100 M in NH, Cl(aq).ApH0.026...

Question

Calculate the change in pH when 3.00 mL of 0.100 M HCI(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH;C (aq) and 0.100 M in NH, Cl(aq).ApH0.026Incorrect

Calculate the change in pH when 3.00 mL of 0.100 M HCI(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH;C (aq) and 0.100 M in NH, Cl(aq). ApH 0.026 Incorrect



Answers

What will be the pH change when $20.0 \mathrm{mL}$ of $0.100 \mathrm{M}$ NaOH is added to 80.0 mL of a buffer solution consisting of $0.169 \mathrm{M} \mathrm{NH}_{3}$ and $0.183 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl} ?$

We would begin this problem by using the Henderson Hasselbach equation to calculate the pH. We can disregard this term, since we know that the concentrations are in a 1 to 1 ratio with each other, so the log of that will just be zero. So we're adding zero to R P K A. The PKK and question was 4.75 In part A. We're adding 0.8 moles of sodium hydroxide to the buffer. This equates to a concentration of 08 Moeller sodium hydroxide. This is a strong race, and it will react completely with the acid itself to form the conjugate base, which means that if we begin with one Moeller acetic acid, this will decrease to 0.92 Moeller. Unlike wise, if we begin with one Moeller of the acetate ion, this will increase 1.8 Mueller. Now we can use the Henderson Hasselbach equation again. The pH, in part a, comes out to be 4.82 Now, In Part B, we're adding 0.12 moles of hydrochloric acid, which is a concentration of 0.12 Moeller. This is a strong acid that will react completely with the conjugate an ion. So the initial one Moeller of the acetate ion will decrease 2.88 Mueller in the initial one. Moeller of acetic acid will increase to 1.12 Moeller again we can use the Henderson Hasselbach equation. The pH comes out to be 4.65

Okay, So the first step in our problem is we need to find the k A of hydrogen fluoride, and this is a known value, and it's 3.5 times tend to negative four. But for the purposes of this problem, we will need the p k k of hydrogen fluoride. It's all of the PKK is what it will take. The negative log of K A. So in this case will take the negative log of 3.5 times 10 to the negative four, and we'll get the value 3.46 So throughout this problem will be using what's called the Henderson Hasselbach equation. And we can find the pH by knowing our p k a and the the concentrations of a base or a con ticket base or an acid, things like that. So, for our problems will be finding the pH, which is equal to 3.46 plus the log of 0.5 divided by 0.25 And when we do this within this, plug it into a calculator, it'll get 3.76 is our Ph. So for the first part of our problem, part A is the first step is we will need to find the molar ity of H and 03 So we take our molds that were given and divided by the volume of our buffer solution. And we will get 0.2 polar. So now we'll, we will do is we will set up was called on a stable. I stands for initial change than equilibrium. So our reaction will be hydrogen plus fluoride which is creating hydrogen fluoride and they're all acquiesced. So for our hydrogen category, we'll start with 0.2 We will be using up all of this and we will start. We will end with zero for fluoride. We will start with 0.5 and we will again use 0.2 up and we'll be left with point for eight of fluoride for hydrogen fluoride will start with 0.25 We will actually gain 0.2 because we're making a product, you're regaining it. We'll end with 0.27 So again we will use the Henderson HASA Bach equation. The exact same way we set it up is before we will get that are pH is equivalent to r. P. K. A. 3.46 plus the log of point for eight, divided by 0.27 and again just spoke this into the calculator. And when we do that, we get that are ph for this first part is 3.71 on to part B, as we will take the same steps as we did for part a. But now just for a base. So we'll find the molar ity of K o h. We'll take. The moles were given divide are the volume of our solution and we'll get 0.4 Moeller against her. But I stable with our reaction. So we'll have hydroxide plus hydrogen fluoride going to fluoride and liquid water. So, for hydroxide, we start with 0.4 We will be using all of that up in this reaction and we'll end with zero hydroxide for hydrogen. Fluoride will start with 0.25 We will again use up 0.4 and we'll be left with point to one for fluoride. We will. These will be starting with 0.5 and you will actually be gaining 0.4 I'm sorry. This should be a plus sign not a minus sign and we will be ending with point 54 and then remembered ice tables. We don't include liquids so we don't have anything in our water category. So again, we will be using our Henderson Hasselbach equation just as we did before. We will get that r P H is equal to a peak a 3.46 plus the law of 0.54 divided by 0.21 again about upon plugging this and we will get there. PH is 3.87 So I hope that these steps help. Do you understand how to use Henderson HASA Bach equation and how to understand what a buffer solution is?

Okay for 18 8 It's also buffer solution so used to Henderson Hassle Baulch equation decay over the base concentration my base concentration over the acid concentration. In order to calculate a change in th, we need to know the initial pH. The initial pH can be calculated looking at the K a value for ammonia and then using the concentrations that have been given to US 0.1 Moeller for the weak base, which is F minus and 0.5 Moeller for the week. Acid, which is Hey and we get 0.73 point 76 22 significant figures in the decimal place, then to solve for the change in Ph. After knowing the initial pH, which would just calculated, we need to determine what the moles of weak acid are that is left. Oh yeah, that is in solution. After we have added the strong acid, the strong acid is going to react. The strong acid, a gentle three is going to react with the weak base F minus and be consumed so it's on mold will decrease the moles of the weak acid. H f will increase both the decrease of the weak base and the increase of the weak acid will be exactly equal to the molds and strong acid added. So the molds of H F that we have initially is the polarity times. The concentration, plus the mole's strong acid, added the moles of weak base. That we have will be the moles of weak base. We start with initial concentration multiplied by volume minus the most of strong acid added, we can then take these moral values and add them into the Henderson hassle bolt equation again and solved for the new P H after the addition of the strong acid. Now, remember, with the Henderson acid vault equation, we can do a ratio of polarities, or ratio of moles. So I'm just gonna put in mole's right here. I put in polarities the volumes of the same so the volumes are gonna cancel, which he left with most anyway, for this, then becomes the new pH. After the addition of the strong acid, as we expected, the pH decreased acids will decrease the pH. So the change in Ph is going to be the final th minus the initial pH, or negative 0.24 molars in order to know whether or not this buffer solution has greater buffer capacity than what was found in problem. Seven. Well, in problem seven, we had a 70.25 Moeller. Hey, Jeff okay, was equal to $0.25 while that minus was equal to point 50 Moeller and you'll notice that the base concentration is twice the acid concentration and we typically have greater buffer capacity when the concentrations are greater or when P H is equal to P. K. Because these two concentrations are greater than the initial concentrations for this problem, then 15.7 because it has higher concentrations of buffering species, has greater buffer capacity.

Answer this question, and we need to recognize that it is a buffer solution so we can use Anderson Hassle Baulch equation P h equals P K A. Plus the log of the base, which is ammonia over the acid concentration, which is ammonium. The ammonium concentration is given to us first as mold of ammonia is and dissolved in 0.5 leaders, so they get the concentration. We divide those two numbers. The ammonia concentration is given to us at 0.4 Moeller. In order to get our peak a value here, we need to recognize that it is the negative log of the K a value who were not given the k a value. But we could look up the cape, be value for ammonia, and you remember the K A value is simply k w divided by K B. And maybe we can look up for the base, which is a moment, often times If we have a predominant, weak base, we will have KP values that we can get K values from the K. A value, of course, corresponds to the conjure. Get acid of that weak base. Then, by dividing those two numbers and taking the negative love we get 9.255 If you carry all of our significant figures back into Henderson Hassle Baulch equation, then we get P H is equal to p K, plus the log of the base concentration or the acid concentration and correctly too. The two significant figures in the Mantis A which is to the right of the decimal place we get 9.56


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