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8 ="ydpka=PK =OuonsALLLL(mL)Figure 10-11 Calculated curves snowin? the titration 50.0 mL of 0.020 M HA with 0.100 M NaOH: the acid becomes weaker; the equivale...

Question

8 ="ydpka=PK =OuonsALLLL(mL)Figure 10-11 Calculated curves snowin? the titration 50.0 mL of 0.020 M HA with 0.100 M NaOH: the acid becomes weaker; the equivalence point becomes less distinct:

8 ="yd pka= PK = Ouons ALLLL (mL) Figure 10-11 Calculated curves snowin? the titration 50.0 mL of 0.020 M HA with 0.100 M NaOH: the acid becomes weaker; the equivalence point becomes less distinct:



Answers

A weak monoprotic acid is titrated with 0.100 $\mathrm{M}$ NaOH. It requires 50.0 $\mathrm{mL}$ of the NaOH solution to reach the equivalence point. After 25.0 $\mathrm{mL}$ of base is added, the pH of the solution is $3.62 .$ Estimate the $\mathrm{p} K_{a}$ of the weak acid.

Questions 15 points 16 is a rather lengthy problem. Has three parts three different type Trish in calculations. In addition to also calculating the equivalence point value which is necessary in order to make sense out of the three additional penetration volumes to give you to solve for the equivalence point volume, you will take the initial volume of the analyze that you have multiplied by its concentration to figure out the moles of the h o. C. L that you have then recognized that the reaction of HCL with an a O. H is a 1 to 1 molar reaction. You will then have at this point the molds of strong base that you need to add using the polarity of the strong base. You can calculate the moles of strong base needed, man using the polarity, you could calculate the leaders of strong base needed. Then you can multiply by 1000 in order to get the volume milliliters of strong base needed for the equivalence point. Volume is 40.0 no leaders. So after the addition of 10 millilitres, we know that we are pre equivalents because 40 is the equivalence when we are pre equivalents with a weak acid being traded with a strong base. We have a solution. So we need to know the moles of weak acid left in solution and the moles of base that we perform. Moles of weak acid left in solution is going to be equal to the molds of weak acid. We started with polarity times volume multiplied by the moles of weak acid that reacted most of weak acid that reacted will be equal to the molds of strong base form. So polarity times following you. Then the molds of week based form will be equal to the moles of strong days. Added for the molds of strong base. Added again is 10 litres multiplied by the concentration that strong based 100.4. Now, knowing the moles of weak acid and it's confident base left in solution, we can use the Henderson Hassle Baulch equation. P H is equal to peek a negative lock into K, a value that was given to us, plus the log of the molds of the base over the most of the acid, which just calculated appear well. Then set Calculate that ph of 6.98 There we go to Part B, which asked us to calculate the pH of the equivalence point pH in the equivalence point for a weak acid strong based filtration which this is always equal to p k p K. Then simply the negative log of Kay You, which was given to us which in this case, a 7.46 Then at the equivalence point, we need to recognize that all of the weak acid has reacted with strong base and has now become weak base. So all we have in solution is the base. If all we have in solution at the equivalence point is the face, then all we have to do is carry out a base pH calculation. A week based pH calculation requires us to no concentration of the weak base. The concentration of the base is going to be equal to the molds of strong base added, or it's also equal to the molds of weak acid. We started with most weak acid we started with is going to be the volume 100 mil leaders multiplied by the concentration 1000.16 You then have to divide that by the new total volume. We started out with 100 mil. Leaders calculated earlier that the equivalence point volume was 40 No leaders. That then gives us a concentration of weak base at 1.14 times 10 to the negative too. The pH From that the hydroxide concentration for a solution containing a weak base is always equal to okay, be value for the weak base which we can get from the K. A value for the weak acid. K B is the kw value which is 1.0 times 10 the negative 14 divided by K. So this right here is our K B value multiplied by the concentration of the weak base and then we square. So that gives us five 5.71 times 10 to make it to five Moeller concentration for hydroxide get pH from hydroxide when you take the negative log of the hydro Liam concentration which we get from the hydroxide concentration by dividing it into kw. So this right here represents the hydro nian concentration that we will take the negative log in order to get our ph of 9.76

So in this problem, were given to depression. Carbs off too weak as they're separated by a strong base. And we have to answer to questions here. First twenties which acid is more concentrated among this to tradition curbs. So if this is our tradition, carve A and this is our technician, Carby will have to look at the equivalence point off each day. Attrition curb To answer this question as we know that the equivalence point ofthe tradition car was basically the have point of the steeper regions in the patrician carb. So if we take the half point here hour Ah, so ah, this is our deadly very querulous point. And if we just draw a straight line from the B A Texas to the cable inspired will see that Ah, this is occuring at a page off ten. And if we drop is straight line on the follow max is will see that the cumulus point is occuring around forty m. L base. That means after adding forty m. L a base, that's it will reach, they could respond. But in case of this Grove B, we can see that if we draw the happy give allowance point halfway through this, we will see that this will be somewhere here. So from the pH, Texas will have a value off around eight and ten and the volume will be around the volume of the base added equivalents point will be around thirty. So that means that in tradition, curve aid acid is consuming more volume of base to reach the equivalence point. And that means that the acid, which is more concentrated, will conceal more volume base to recycle its point. And therefore, since we have ah more ah, volume of base consumed in carp A. That means that the acid element carb A or their sit in car is more concentrated than that. Sit in car B. No. Ah, in second question, we have to answer, which is it has larger k f l O, and we have to I considered the same thing here. The acid which will consume less volume off base or which will require less volume of base d'Oh our is the government's point will mean that that will be less concentrated and less concentrated acids have larger Kiev fellows. That means, ah, the lower the concentrations ofthe acid, the larger the cave alone. That means since in our car B we have lower volume a base consumed compared to the graf A where we have a higher volume, a base consumed that means our acid in Graff B will have larger gave alot.

For this question. It first asks us to sketch the tray shin curve for the weak acid h A being tight, traded with the strong base in a wage. A weak acid. Strong based Hi Trey Shins. We'll start out with a low pH and then it will increase close to the equivalence point and you will end up with a Ph. D. It would last point, which is halfway up the steepest portions of portion of the curve. Uh, pH. It is going to be greater then, and that's what we have here is a weak base has been created from the castle. Calculate the middle leaders of sodium hydroxide that is required to titrate 50 milliliters. Uh, weak acid. You can carry out a unit conversion knowing that every 1000 no leaders have a leader and the polarity of the weak acid H A is 0.1 So for everyone beater 0.1 Okay, these, like geometry of the reaction is one for everyone more in the wage, and then the polarity of the N E O. H will allow us to calculate the volume of any awaits that's required. It is also there a 0.0 moles in your wage, Peter. And of course, one leader. So to reach the equivalence point, carry out this calculation. We're going to need 0.0 milliliters. Wait part. See? Yeah. Is the pH of the equivalence point greater than equal to or less than seven? And that question waas answered here for part D. What is the Ph exactly halfway to the equivalence. So we're looking at the pH about here. And if you remember pH Ukuleles p k at the half quick glimpse point. So the K a value for this weak acid? Yes, zero times before. So the P K, the negative log of that and simply 4.0

If we have a base being titrate ID with a strong acid, we'll start out at a high pH and end up at a low pH. So for the strong base, being tight traded with HCL is gonna look something like this. We see that the base concentration is twice that of the acid concentration. So the acid volume required is going to be twice that of the base. The base was 50 mL. So we're going to need 100 mL of HCL to reach the equivalence point. And we would have a pH at the equivalent point of about seven for the strong base strong acid titrate shin. For the one at which the KB value is one times 10 to the negative five. It would look something like this where the pH at half equivalent is going to be equal to p k A. Que is going to be one times 10 to the negative nine. So PK is going to be nine and then for the next one where K B is one times 10 to the negative 10. Then K is one times 10 the negative five. And at the half equivalence point, we're going to have a pH of about five, so this would be an example of what the graph would look like.


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