Find the pH of the equivalence point(s) and the volume (mL) of 0.0588 M KOH needed to reach the

Find the pH of the equivalence point(s) and the volume (mL)...

Key Concepts

Polyprotic Acid Titration

Polyprotic acids can donate more than one proton and thus have multiple equivalence points during titration. Each equivalence point corresponds to the neutralization of one proton, and their pH values may vary significantly due to the formation of different intermediate species.

Stoichiometric Calculations

Stoichiometric calculations in titration involve determining the number of moles of the acid or base present, based on its concentration and volume. These calculations are crucial for figuring out the exact volume of titrant required to reach the equivalence point, ensuring accurate quantitative analysis.

Acid-Base Titration

Acid-base titration is an analytical method used to determine the concentration of an acid or a base by gradually adding a titrant of known concentration until the reaction reaches a known endpoint. This process relies on precise measurement and reaction stoichiometry.

Equivalence Point

The equivalence point is the stage in a titration where the amount of titrant added exactly neutralizes the analyte, according to the reaction stoichiometry. It is essential for determining the reaction’s endpoint and for calculating concentrations of substances in solution.

Weak Acid-Strong Base Titration

In a titration of a weak acid with a strong base, the equivalence point typically exhibits a basic pH due to the hydrolysis of the conjugate base formed. This concept is important for understanding and computing the pH at the equivalence point when the acid’s conjugate base affects the final pH.

Transcript

00:01 Here we have a reaction of hno2 with k -oh, which forms water and kno2.

00:08 If we're titrating 23 .4 milliliters of 0 .039 molar hno2 with 0 .0588 molar k -oh, in order to get to the equivalence point, we're going to want to completely react the hno2 without adding any excess k -o -h so that only water and k -n -o -2 are present in the solution.

00:37 So in order to do that, we have to have an equal number of moles of h -n -o -2 and k -o -h, since there is a one -to -one ratio between the two.

00:48 So first, let's calculate how many moles of h -n -o -2 we have.

00:53 Polarity is moles over liters.

00:56 So our number of moles of hno2 is equal to the molarity of hno2, which is 0 .0390, times the number of liters of hno2, which is 0 .024.

01:19 So we know we have 0 .00913 moles of hn02.

01:27 That means we also need 0 .000913 moles of k -oh to completely react with the hno2.

01:42 So now with this information, we can solve for the volume of k -o -h we need to add.

01:49 Using the same equation here, we have liters is equal to the number of moles over the molarity.

01:59 So the number of liters of koh we need is 0 .000913 moles over 0 .088 molar.

02:18 And this is equal to 0 .0155 liters of koh or 15 .5 milliliters of koh.

02:33 So we have the volume of k -o -h needed to reach the equivalence point.

02:42 And now let's find the ph at the equivalence point.

02:47 All that exists in solution at the equivalence point is k -n -o -2 and water.

02:53 So let's look at a reaction between these two.

03:15 In solution, these ions will separate from each other, and so will these.

03:20 And so potassium is what's called a spectator ion.

03:24 So we can rewrite this equation without potassium.

03:27 Now let's set up an ice table.

03:48 Our first step is to figure out how much, or what the concentration of no2 minus is in the initial step.

03:59 We don't have to worry about water since the liquid, and we don't use the liquids in our calculations.

04:09 So we know that in order to reach the equivalence point, we reacted 0 .000913 moles of hno2 with the same 9 -02.

04:19 Number of moles of k -oh.

04:22 Since there is a one -to -one ratio of each of these with no2 minus, we know that there will be the same amount of moles of the no2 minus at the equivalence point.

04:33 So we know we have 0 .00913 moles of no2 minus at the equivalent's point.

04:43 Now in terms of volume, we started with 23 .4 milliliters of h -n02, and we added 15 .5 milliliters of k -oh.

04:56 If we add those and convert to liters, we get a total volume of 0 .389 liters.

05:07 This then gives us a molarity of 0 .025 molar n02 minus.

05:17 So now we finally have our initial concentration of n02 minus, which can go in our ice table here.

05:25 And our initial concentrations for the products are 0 molar for each of them.

05:32 Our change is then some unknown x.

05:35 So we react x, n02 minus, and we form x of each of the products.

05:44 Then our equilibrium is just our initial plus our change.

05:49 So for n02 minus 0 .035 minus, for each of the products, just x.

06:00 Now we know that the kb of no2 minus is equal to the concentration of hno2 times the concentration of oh minus over the concentration of no2 minus, and we also know that kw is equal to ka times kb.

06:25 This is important because we can look up the k -a value of no2 minuses conjugate acid, which is h -n -o -2.

06:37 We can look that up in the textbook, and the value of that is 7 .1 times 10 to the negative 4th.

06:47 So we can use this number here to convert to kb, which we can then use in this equation here to solve for our values of x.

06:59 So let's do that.

07:02 Plugging in our ka into this equation here, we get a kb of 1 .41 times 10 to the negative 11th.

07:14 So 1 .41 times 10 to the negative 11th is equal to our concentration of hno2, which is x in our table, times our concentration of oh minus, which is also x.

07:29 So i can just write this as x squared, divided by our concentration of no2 minus, which is 0 .035 minus x.

07:45 In order to make solving for x a little bit easier, let's make the assumption that x is so small that subtracting it from 0 .035 here is not going to make a difference.

07:56 We can make this assumption as long as we check it at the end to make sure it was valid.

08:02 So if we make that assumption, then we have 1 .41 times 10 to the negative 11th is equal to x squared over 0 .0235.

08:17 Solving for x, we get that x is equal to 5 .75 times 10 to the negative 7th.

08:26 And this tells us that our assumption is valid since 0 .0 .235 minus 5 .75 times 10 to the negative 7th is still 0 .0235.

08:37 Five to three significant figures.

08:41 If our assumption was invalid, we would have to go back and re -solve this problem for x using the quadratic equation.

08:51 But since our assumption was valid, we can just move on.

08:55 Now that we have our value of x, we also have our value of the concentration of hydroxide at our equivalence point.

09:06 That's because at our equivalence, at our equivalence, at our of equilibrium, we have x molar oh -h -minus.

09:18 And now we finally have enough information to solve for a ph.

09:24 Ph is equal to 14 minus p -oh -h, or 14 minus the negative log of the concentration of oh -h -minus.

09:38 Plug the in our concentration of hydroxide, which is equal to x, we get a final ph of the of 7 .76.

09:50 Here we're titrating h2 -c -o -3 with k -o -h will result in two equivalence points because h2 -c -o -3 has two protons to lose.

10:02 First it will form h -c -o -3 minus, which will then further react with k -o -h to form co3 -2 -minus.

10:15 So let's look at these reactions one at a time.

10:19 First we start with 17 .3 milliliters of 0 .1 .1.

10:24 103 molar h2 co3 and 0 .0588 molar k -o -h.

10:41 Just like in the last problem, we can multiply the molarity by the volume to get the number of moles.

10:50 So if we multiply 0 .130 by 0 .073, we get our number of moles of h2 -co3, which is 0 .00225.

11:15 Now we know we need 0 .0025 moles of k -oh as well, since we have a one -to -one ratio of k -oh and h2 -c -o3 that react in our titration...

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