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REFORT SHEET Determination of the Ka for Weak Acid Nate Date Section _ Partner' $ Name Hnetructor Hleeeua cht nt "I6 IL NaOH Molarity OLOKMar) Eflicicncy ...

Question

REFORT SHEET Determination of the Ka for Weak Acid Nate Date Section _ Partner' $ Name Hnetructor Hleeeua cht nt "I6 IL NaOH Molarity OLOKMar) Eflicicncy of pH mcter (15) Since the titralion with A-mL increments done give Tough Csumatc the equivalence point; you nced report on only the results trom Ulte more precise increment titrulion:Acelic Acid Literalure value for Ka of Acetic AcidLilcrnire Valuc for pKa of Acetic AcidVolume ofbasc reach equivalence pointequivalence pointVolume of

REFORT SHEET Determination of the Ka for Weak Acid Nate Date Section _ Partner' $ Name Hnetructor Hleeeua cht nt "I6 IL NaOH Molarity OLOKMar) Eflicicncy of pH mcter (15) Since the titralion with A-mL increments done give Tough Csumatc the equivalence point; you nced report on only the results trom Ulte more precise increment titrulion: Acelic Acid Literalure value for Ka of Acetic Acid Lilcrnire Valuc for pKa of Acetic Acid Volume ofbasc reach equivalence point equivalence point Volume of bast halfwvay eqquivalence point halfivay cquivalence point Experimental pK. of Acetic Acid Pereent etor of pK, Experimental Ka Acetic Acid Penenenar ofk. Show both percent emor calculations (4pLk: Unknown code Volume of base Teach equivalence point equivalence point Volume ofbase halfway equivalence point



Answers

Consider the titration of a 20.0-mL sample of 0.105 M HC2H3O2
with 0.125 M NaOH. Determine each quantity.
a. the initial pH
b. the volume of added base required to reach the equivalence point
c. the pH at 5.0 mL of added base
d. the pH at one-half of the equivalence point
e. the pH at the equivalence point
f. the pH after adding 5.0 mL of base beyond the equivalence point

So now I work on problem on twenty from Chapter seventeen here we're told that we have ah, mass sample of a weak acid with a K of one point three times ten to the minus four combined with a volume of sodium hydroxide. And the solution is that I looted and the measured pH of the solution is four point two five, and it asks us for the Mueller Mass. So first one scan, we can calculate the base of acid that were actually adding here, the moles of base molds of basic we're adding. And so we get that by multi plane concentration times volume to get moles. So next we're told. PH it's four point two five, we're told the K so in calculate P K as being equal to three point eight nine, plus the log of base over essence. Now we're not sure what the concentration of incidents because we don't know the molar mass. But we do know that base the amount of consequent base because it will be equal to this mountain because the sodium hydroxide will react completely with the weak acid. So when we do this calculation, we subtract three point eight nine from both sides and take the anti log. We get that this ratio here's the base to the acid is equal to two point two nine just one two. And so when we divide zero point zero three bye two point two nine, we get the moles of acid, or zero point zero one three moments. So that tells us that the total congregate passive base moles can be calculated and it like. So So we add the ass malls of the acid as well as the malls for country of base. That's equal to zero point zero three from the strong base that we added and also from the calculated acid that we added. And we get a total zero point zero for three moles of a minus and H A. So that's our total. Moles were given the mass of beginning so we can divide grams by moles to get Mueller Mass, and that is equal to one twenty nine point zero seven grams for more

Hello. So today we'll be looking at I Trey, Trey Shin off hydrochloric acid, which is a strong acid, and Mazzella mean, which is a weak base. So what happens when we reached the equivalents Point? Well, at the equivalents point all of this week, base will be converted to the conjugal acid. So what will be the P H at the equivalents point, while everything has been converted to the conjugal acid and the Hakan ticket acid is going to react with water partially to reform the conjugate base and form hydro knee, um, I on. So first, let's try to figure out what the of initial concentration off this weak ass it is while we don't know the volume or the multi moles that we start out with, but we do know the concentration. So say hypothetically that we had one the leader off this weak base metal mean so we had 0.2 moles of it. Well, we would need one leader of hydrochloric acid to react with methyl amine to form the contract acid, so we would have this. So the 0.2 moles of methyl amine would become pope 0.2 moles off the con jacket acid, and we would end up with two leaders as the volume. So we would divide eso the it becomes more dilute. So it would end up with zero 0.10 polarity of this acid. And if you take a look, you'll notice that it doesn't matter what concentration you start off each, you start out with, you will always get that it's going to end up being 0.1 molar ity. And then we have zero of the Mesilla mean and the hydro Nehemiah is about zero because technically, it's 10 to the negative seven. But this is very small Suijk and consider it negligible. So then this acid will react to create one mole of this acid will react to create one mole of the conjugated basin, one mole of hydro knee, um, violence. And so we will see one bullet, one mole arat e 0.1 mole Araji subtracting the change. But this change is very, very small compared to this 0.1. So we will just approximate it as 0.10 molar ity and then the methyl amine in the high droney amiss simply acts. Now the k a for this acid is 2.3 times 10 to negative 11. And the K is the conjugate base, which is Mesilla mean times the hi dhoni, um, over the acid. So you'll notice that it's equal to X squared over 0.1. And so, if you just rearrange this, you'll get X squared is equal to 2.3 times 10 to the negative 12. So we simply take the square root of both sides, and we will see that X is equal to 1.5 times to tend to the negative six. Similarity. So now this is equal to the hydro knee, Um, concentration. So if we finally want to find out the pH. The pH is the negative log with the hydro knee, um, concentration, which is the negative log of 1.5 times 10 to negative six, So the pH is five 0.8.

In this problem, we're told that we have a weak mono protic acid and it's tight traded with zero 100 Moller, strong beats starting my drug side. And it requires 250 ml to reach the equivalent of point After 12.5 mm of the base have been added. The Ph is 4.16. We are asked to estimate the peka Well 12.5. Is that halfway point between for the tight rations and at the halfway point at the half equivalence point, the peak A. Is equal to the ph. So the PKA will equal 4.16 easiest problem.


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