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1.12 L buffer solution is 0.142 M in CH3COOH and 0.267 M inNaCH3CO2. What is the pH of the solution after adding 19.2 g ofNaOH (s). Ka of CH3COOH=1.8*10^-5...

Question

1.12 L buffer solution is 0.142 M in CH3COOH and 0.267 M inNaCH3CO2. What is the pH of the solution after adding 19.2 g ofNaOH (s). Ka of CH3COOH=1.8*10^-5

1.12 L buffer solution is 0.142 M in CH3COOH and 0.267 M in NaCH3CO2. What is the pH of the solution after adding 19.2 g of NaOH (s). Ka of CH3COOH=1.8*10^-5



Answers

A buffer containing 0.2000$M$ of acid, HA, and 0.1500$M$ of its conjugate base, $A^{-},$ has a pH of $3.35 .$ What is the pH after 0.0015 mol of $\mathrm{NaOH}$ is added to 0.5000 $\mathrm{L}$ of this solution?

If we start with a buffer solution that contains equal concentrations, uh, di hydrogen phosphate and hydrogen prostate, and the pH will be equal to PK, the PK of the acid that's president, which is di hydrogen phosphate. So the pH would be 7.2 if we then add some sodium hydroxide. Sodium hydroxide is going to react with the di hydrogen fost eight and in the process, form hydrogen fost eight. So we're gonna increase the moles of hydrogen prostate and decrease the moles of di hydrogen prostate that will adjust the pH accordingly, so we'll still use the Henderson Hassle Baulch equation. PH equals P K, plus the log of moles and face we start with, plus the moles of strong base that were adding because every mole of strong base, we add will create a mole of wheat based. And then in the denominator, the moles of asked if we start with two leaders multiplied by the concentration will give us the moles of hydrogen phosphate or sorry di hydrogen phosphate, which is the acid. It's gonna react based on the molds of sodium hydroxide added so minus the mold of sodium hydroxide, which is represented by the massive sodium hydroxide divided by its lower mass, and we get a pH of 7.24

Using the hand. It's an equation we can deter mined the pH off acid solution or based solution. Well, sorry for that. We have the K that is the association value for asset equivalent to 7.1 independent super minus tree. There's a relationship between P, K and K, as shown to calculate the peek a value we subsidy the key into the given equation, which gives us a peek a equally lento 0.15 Calculate the concentration off 0.155 mole. 1/4 uric acid in 500 ml off water shows the concentration, which is 0.310 Calculate the concentration off 0.250 Cool off kids to be or four and wind 500 leader off water, which is also 500 ml. The concentration gives us 0.500 m. The concentration off phosphoric acid is 0.310 on the concentration off kids, too, p 04 turns out to be 0.500 To calculate the pH of solution we substitute the values into the hand is an equation. Substituting these value in tow equation to deter mined the Ph, which comes out to be 2.36

So here we're give an example of a weak acid conjugate based solution. So we can use our Henderson Hasselbach equations, that ph equal to the PKK put the loss of concentration of our conjugate base over the weak acid. So we're given information that our initial ph is equivalent to 8.77 And our concentration of our weak acid is equivalent to 0.110 molars. And our concentration of our weak base, Our continent basis equivalent to 0.220 molars. And we can use this information to find the PKK which we're going to need later. Now substituting everything in. Okay? Mhm. We can find that the peka is equivalent to eight points 8.469. Paying attention to significant figures. Only the log digits after the digits after the logs are significant. So in this case we have our PK here. Now we're going to stress the system and were given that the volume volume of the solution Is equivalent to 0.350 leaders. And we're going to stress the system by adding a strong base in terms of hydroxide. So the only compound that can react with a strong base in this case is our weak acid. So h Y would react with hydroxide to yield. In this case it would actually be a complete reaction. Since there is a significant difference in the case between the two that this would yield water plus the conjugate base. And we can see that the concentration of conjugate base will increase while the concentration of the week asked will decrease. So we have to account for those calculations. So the number of moles of our weak acid is equivalent to final would be equivalent to the initial number of moles, which is just concentration times volume according to the concentration formula. Then we're subtracting in this case, since we're using barium hydroxide, There are two moles of hydroxide contributed per mole barium. So the concentrations, the concentration effect is doubled. So we are adding 0.0015 moles. Uh barium hydroxide so that the concentration effect would be doubled. So 0.11 times .35. Then we're subtracting two times 0.0015. And we can see that our final number of moles is 0.0355 moles. Okay. And our mole number of moles of why minus prime is equivalent to our initial. And we're using in addition sign in this case, since there is an increase in concentration times two times 0015 most. Since all of that you consumed age wise, used to make the counterfeit base. So now we can substitute everything in 0.2, two times .35 Plus two times 0.0015. This is equivalent to 0.0800 moles. And in this case we don't have to use concentration. Since when we express this concentrations, the volumes were just cancel out so we can ignore the concentrations in this case and just express it as moles. So ph equals P. K. Was logged of the concentration of conjugate base modified over the concentration of weak acid modified. And let's remember that our PA. 8.469. Now we substitute our values. And yeah. And we find that the PPP becomes eight points 8 to 2. Which makes sense considering we're adding strong base too, a strong base to a solution which would obviously contribute to raise the ph in this case. And that's it.

Saying weak acid H Y partially associates into. I drove him. I owns and why my in this forms a buffer. If this buffer has a pH of 8.77 where H y has a concentration of 0.110 Moeller and my minus has a concentration of 0.220 more, we can calculate the peak A of H white using the hundreds and how Slovak Equation Rearranging this we get P K eight equals ph minus the log of the buffer component concentration ratio. You know we can clog undervalues here. This isn't an equal to 8.77 so minus 0.3 or one, which is 8.47 So don't let's say we add 0.15 moles. Oh, very and hydroxide to to your 0.350 leaders of the buffer solution. We can find the concentration of barium hydroxide since polarity is just most of our leaders. So it's your point. Syria 015 nos over 0.350 liters. Then we get the report 004 to 9 Moller barium hydroxide. This means that we have 0.4 to 9 Miller caring and twice as much. So zero point years You're 857 Moeller hydroxide. This is because there is a 1 to 1 ratio between very hydroxide and burial and 1 to 2 ratio of barium hydroxide to buy drugs. The weak acid H wine is going to react with the hydroxide plans from the added barium hydroxide to create water and one minutes hydroxide is a strong base. So this reaction is gonna go to completion, which is why there is ah, single Ford facing arrow. So now we consider open ice table where actually we have 0.110 Moeller h y 0.8 00857 Moeller o H minus which is this number here and 0.220 Moeller y minus. Liquids are included in these calculations, so we can just ignore water here, someone to find the change. We know that Oh, um of the reactions on the left side that can rear will. And since hydroxide is the limiting reagent, we know all of it Well rare. So the change for hydroxide is negative. 0.857 Moeller and since there is a 1 to 1 ratio between hydroxide and h y, but the same concentration of H Y is going to react as, um, always minus. So the change for H wise also negative. 0.857 Moeller. Then again, since there was a 1 to 1 ratio between hydroxide and what why minus, um, the amount of hydroxide that react will form the same amount of Y minus on the other side, so the change for more minus is a positive. The record 00 857 Moeller. Then for equilibrium or end, we just add initial and change. So for h y, we have so 0.10143 fallen for a dockside. We end with zero Moeller number y minus. We end with 0.21143 Moeller. And now that we have these final concentrations for y minus and h wife, we can use them to calculate the peach of the final solution using the hundreds and Hasselbach equation again so we can plug in our calculated value of PK since that won't change. And then we complain when our new co concentrations that we found in the ice table. So 0.21143 ever. It's your appoint 10143 This is a 24 7 close, 0.319 so or 8.79 and that's a final Ph.


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