EXPLANATION OF ACID - BASE PROBLEMS


 

TITRATION AND pH PROBLEMS

THE FOLLOWING SALTS IN WATER SOLUTION WILL HAVE A pH OF 7, -7(less than 7), +7(more than 7), OR INDETERMINATE (I). FOR EACH PREDICT THE pH OF THE SOLUTION.

1. Na2CO3 is a basic salt because NaOH is a strong base and H2CO3 is a weak acid. pH is above 7.

2. FeCl3 is an acid salt because Fe(OH)3 is weak base and HCl is a strong acid. pH is less than 7.

3. KNO3 is a neutral salt. Both HNO3 and KOH are strong. pH = 7.

4. NH4C2H3O2 is indeterminate. Both NH4OH and HC2H3O2 are weak. pH is actually only a very little above 7.

5. ZnSO4 is an acid salt. The first ionization of H2SO4 is strong and Zn(OH)2 is weak. pH is less than 7.

6. Ba(NO3)2 is neutral because both Ba(OH)2 and HNO3 are strong. pH = 7.

7. RbF is an alkaline salt because RbOH is strong and HF is weak. pH is over 7.

8. CaBr2 is neutral because both Ca(OH)2 and HBr are strong. pH = 7.

Back to the Acid and Base math problems.

9. What is the pH of a 0.0115 M HCl solution?

The only two materials in the solution are water and HCl. The water alone would provide 1 E-7 Molar hydrogen ion. The HCl is a strong and soluble acid, so all of the acid is in the form of ions, providing 0.0115 Molar hydrogen ion. There is no need to consider the hydrogen ion from water as it is much less than five percent of the total hydrogen ion concentration. The concentration of hydrogen ion is adequately estimated at 0.0115 M. This is only a pH box problem. (Note the symbol Chemtutor uses for pH box math.) As a final answer, Chemtutor suggests rounding pH values to one decimal place. Few pH meters are that accurate.

Back to Acid and Base math problem #9.


10. Find the pH of 0.0815 M NaOH solution.

The only two materials in the solution are water and NaOH. The water alone would provide 1 E-7 Molar hydroxide ion. The NaOH is a strong and soluble base, so all of the base is in the form of ions, providing 0.0815 Molar hydroxide ion. The hydroxide ion is most important in this problem. The hydroxide ion from the dissociation of water is much less than five percent of the hydroxide ion from sodium hydroxide, so the concentration of hydroxide ion is 0.0815 M. This is only a pH box problem. The answer has been rounded to one decimal place.

Back to Acid and Base math problem #10.


11. Find the pH of 0.00372 M Ba(OH)2 solution.

Ba(OH)2 is a strong base, that is, up to its solubility limits the compound will be completely ionized in water solution. [(OH)-] is twice the [Ba(OH)2] because for each mol of Ba(OH)2 there are two mols of hydroxide ion. The ionization of water, the only other possible source of hydroxide ions, produces nowhere near five percent of the hydroxide ion available from Ba(OH)2, and so:

[(OH)-] = 2 [Ba(OH)2] = 2 (0.00372 M) = 0.00744 M
Now that we know the [(OH)-], this is only a pH box problem. The answer is in pH, so it is rounded to one decimal place.

Back to Acid and Base math problem #11.


12. Find the pH of 0.12 M HC2H3O2

The only two things in this solution are water and acetic acid, a weak acid. The kA of acetic acid is 1.74 E-5. Water and acetic acid could contribute hydrogen ion to this mixture. If water were the only possible contributor, it would give 1 E-7 M hydrogen ion. If the acetic acid were the only contributor, we could calculate it from the equilibrium expression. Since the hydrogen ion from the acetic acid would be exactly the same concentration as the acetate ion, [H+] = [(C2H3O2)-]. And we can simplify by substituting [H+]2 for [H+] [(C2H3O2)-].

Now we can substitute our numbers for the kA and [HC2H3O2] to find that [H+] = the square root of (0.12 x 1.74 E-5). The [H+] due to acetic acid is 1.4450 E-3 M.

The [H+] from the acetic acid is less than five percent of the [acetic acid], so there is no need to get a more accurate answer by subtracting the [H+] from the [acetic acid]. After finding the [H+], we calculate the pH by using the pH box. The pH is 2.840134, rounded to 2.8.

Back to Acid and Base math problem #12.


13. Calculate the pH of 0.0000135 M H3BO3

Oh, no. We have a complication. H2CO3 has such a small kA (5.37 E-10 for the first ionization) and the acid is in such small concentrations that the [H+] from the acid will be not much more than the [H+] from the ionization of water. The other possible sources of [H+] are the second and third ionizations of the boric acid, but they both have kA's that are far from the first ionization, so they should not be of any significance.

What you would find, if you assume the ionization of water to be negligible, is a silly answer. You try it. Use the standard square root of ( kA times [H3BO3]). You get a pH OVER seven. (pH = 7.1) Ridiculous. Boric acid is an acid and should result in a pH of less than seven.

There are several ways to do this. One way is by successive approximations. First use the ionization constant formula to calculate the [H+] from the acid, then use that [H+] to find the [OH-]. Since the [H+] from water equals the [OH-], you can add that number back to the [H+] from the acid to get a first round total [H+]. Use the first round total [H+] in the acid dissociation equation to find the concentration of acetate ion and that would be equal to the second round estimation of the [H+] from acid. Keep going around that circle until you get three stable significant digits for the total [H+].

Another way is by substitutions into known equations. Your equations are the formula for the kw, the formula for the [H+] of a weak acid, the equality of the [H+] from water with the [OH-], and the statement that the total [H+] equals the [H+] from water plus the [H+] from the ionization of the acid.

Or you could substitute in for the [OH-] ion and solve for the [H+].

Drop the numbers in and do the math. It is messy and tedious and disgusting to have to use the quadratic equation, but it will give you and answer of [OH-] = 6.6115 E-8 which results in a pH of 6.8203, rounded to 6.8, a likely answer.

Back to Acid and Base math problem #13.


14. pH of 0.255 M NH4OH

The kB of NH4OH is 1.78 E-5.

The ionization equilibrium expression for NH4OH or any other weak base is similar to the ionization equilibrium expression for a weak acid, but this time the equation will be solved for the hydroxide ion concentration.

The hydroxide ion is in excess. The two sources of hydroxide ion are the [H+] and water, but the contribution from the water is negligible.

The other assumption, that the amount of dissociated ion is negligible, is also valid because the concentration of base is so large. (Try it if you don't believe it.)

Now substitute the numbers and do the math.

Back to Acid and Base math problem #14.


15. 0.578 M H3PO4

In a solution that has only water and phosphoric acid there are four possible sources of hydrogen ion, the water and the three ionizations of the phosphoric acid. Each ionization has a different kA.

H3PO4 ===> (H2PO4)- + H+ first ionization kA = 6.92 E-3
(- ===> (HPO4)2- + H+ second ionizaton kA = 6.17 E-8
(HPO4)2- ===> (PO4)3- + H+ third ionization kA = 2.09 E-12

It is clear that the first ionization of phosphoric acid at this concentration is much greater than the third ionization of phosphoric acid, but the big questions are whether we need to consider the second ionization of phosphoric acid or the ionization of water.

The hydrogen ion concentration from the first ionization can be calculated by the usual square root of ([acid]kA) to get an initial estimate.

From the initial estimate, the [H+] from the first ionization (0.063243656 M) will be more than 5% of the acid concentration (0.578 M). We will have to use the more accurate form of the equation that includes the decrease in unionized acid concentration by the ionized forms.

Will we have to also account for the second ionization? The number we found as first estimate for the [H+] can also serve as estimate of the [H2PO4)-], the anion in the first ionization. But the anion of the first ionization is the original acid in the second ionization and the estimate of the [H+] can also be used in the second ionization equation. If we solve for the anion of the second ionization equation, we would get an estimate of the hydrogen ion that the second equation would produce.

This is an interesting result. Notice the [H+] from the second ionization is very close to the second kA when there is nothing but water and phosphoric acid in the solution. What would happen at other concentrations of phosphoric acid? At other pH's whith the same initial concentration of phosphoric acid?

The concentration estimate of the hydrogen ion from the second ionization is far below 5% of the hydrogen ion from the first ionization, so the second ionization does not need to be considered. But we do need to consider the decrease in acid concentration from ionization of the first reaction, so we use the quadratic form of the equation substituted into the quadratic equation.

When you substitute the numbers, the math looks like:

As you can see when you do the math, the [H+] is 0.059878231, more than 5% different from the original estimate. Now we can confidently change that [H+] to the pH by the pH box.

If you are really sharp, you noticed that the original estimate using the simplified method of calculating the [H+] produces the number 0.063243656, and that the pH of that [H+] is the SAME as the number from the more complex calculation when rounded to one decimal point.

Back to Acid and Base math problem #15.


16. What is the pH of 0.16 M HCl and 0.072 M phosphoric acid?

There are five sources of hydrogen ion in this solution. Count them. Water, HCl, and the three ionizations of the phosphoric acid. Likely the water and the lesser two of the three ionizations of phosphoric acid will not significantly contribute to the hydrogen ion concentration.

Since the HCL is a strong acid, it contributes 0.16 Molar hydrogen ion to the solution, but the phosphoric acid is a weak acid and the [H+] from it can not calculated by the square root of (kA [H3PO4]) because the [H+] is not equal to the [H2PO4]. We must go back to the ionization equilibrium equation for the first ionization of phosphoric acid.

We can look up the kA. The [H+] is the same as the [HCl], and the concentration of phosphoric acid is given. The [H2PO4] will be equal to the concentration of hydrogen ion contributed by the first ionization of the phosphoric acid.

So the pH depends only on the [HCl].

Back to Acid and Base math problem #16.


17. Find the pH of 1.25 M acetic acid and 0.75 M potassium acetate.

Acetic acid kA = 1.74 E-5       pKA = 4.76.

This is a genuine buffer problem. Added to the water are a weak acid and a salt containing the anion of the acid.

There are two good ways to work buffer problems, with the Henderson- Hasselbach equation or with the ionization equilibrium expression of the weak acid or base. I personally have a mental block against the H-H equation because I can never remember whether it uses a positive or negative log and which concentration goes on top. You can use it if you wish. Particularly if you need to calculate buffers often, you should engrave it upon your gray matter. If you really need it and can't remember it, you can derive it from the ionization equlilbrium expression.

There are three cautions you need to observe with either equation: (1) Make sure you are using the correct concentration for each variable, (2) check to see if the numbers you propose to use are going to be within the 5% rule for simplification, and (3) estimate the answer from what you know and make sure your final answer is reasonable.

Before actually doing the problem, estimate the answer from your own reasoning. In this case, the pkA of acetic acid is 4.76. The rule is that an equimolar buffer has a pH equal to the pkA and in this problem there is less potassium acetate than acetic acid, so the pH must be lower (more acid) than the pkA within a pH unit or so. If the acetic acid were the only solute, the pH estimate would be the square root of (acid concentration times kA).

The answer should be somewhere between pH of 2.3 and 4.8

The majority of the acetate ion will be from the potassium acetate. Is it right that the total acetate ion concentration will be equal to the concentration of the potassium acetate? Or will the acetate ion concentration from the ionization of the acetic acid contribute more than 5%? The potassium acetate concentration is 0.75 M. The acetate ion concentration from acetic acid would be 0.00466 M, less than 5% of 0.75 M even without the common ion effect. We can safely use 0.75 as the concentration of acetate ion.

Will the concentration of unionized acid be a problem? The measured concentration is 1.25 M and the ionized amount is 0.00466 M, far less than 5% of 1.25 M.

As threatened, we can use the ionization equilibrium expression of acetic acid for the main equation for this problem, substituting for the kA, substituting the concentration of potassium acetate for the concentration of acetate, substituting the concentration of acetic acid, and solving for the hydrogen ion concentration to get the pH.

The answer of pH = 4.5 is a reasonable one by our estimation because it is more acid than the pkA of 4.76.

It is a little easier to do this problem by the Henderson- Hasselbach equation, if you are sure you know it. You must still make sure you are substituting correctly and that your assumptions for simplification are valid (within 5%). The H-H equation is not much good for solutions in which either the acid or ion concentrations are more than ten times one another or in which the concentration of either material is less than one hundred times the kA because it doesn't easily adapt to a quadratic form.

Back to Acid and Base math problem #17.


18. 0.788 M lactic acid and 1.27 M calcium lactate.

Lactic acid kA = 8.32 E-4       pKA = 3.08.

Here is another acid - conjugate base buffer pair. This time there is more conjugate anion than acid concentration, so we expect the pH to be somewhat higher (more alkali) than the pKA. As in the previous problem, there seems to be no complication with either of the components being of too small a concentration or the concentrations being too close to the KA, so there should be no need for a quadratic equation.

The concentration of calcium lactate needs to be doubled (!) to represent the lactate ion concentration because the calcium is divalent and has two lactate ions per formula of calcium lactate. The concenration of acid is more than 100 times the KA, so the concentration of acid is close enough to the concentration of unionized species.

By the ionization equilibrium equation:

Or by the H-H equation, you get the same answer.

Back to Acid and Base math problem #18.


19. 0.590 M ammonium hydroxide and 1.57 M ammonium chloride.

ammonium hydroxide kB = 1.78 E-5       pKB = 4.75.

Here we have a weak base and its conjugate cation. We can use the ionization equilibrium expression, but it is different from the acid ionization expression. The ammonium hydroxide ionizes into hydroxide ion and ammonium ion, so it would be best to find the concentration of the hydroxide ion.

The ionization equilibrium expression must have the kB rather than a kA, or the Henderson- Hasselbach equation has to have all its components adapted to alkali, but it is completely analagous to the acid calculation. In either way of doing the problem, you will have to change the answer to the pH by the pH box.

Will we be able to use our standard shortcuts? The concentration of base is more than 100 times the kA, so the measured amount of ammonium hydroxide in solution is a good enough number for the concentration of unionized species. The concentration of weak base and conjugate ion will be within 1:10 of each other, so the amount of conjugate ion can be adequately estimated by the concentration of ammonium chloride. There is high enough concentration of the base so that the ionization of water does not significantly change the hydroxide concentration.

Or by the H-H equation, you get the same answer.

Does the answer make sense? The combination is a base buffer and the pH is slightly base. There is almost three times the concentration of ammonium chloride than ammonium hydroxide, so the pH of the mixture is more acidic than it would be if the buffer had been equimolar. (pH = 9.25)

Back to Acid and Base math problem #19.


20. Explain how to make 5 L of 0.15 M acetic acid-sodium acetate buffer at pH 5.00 if you have 1.00 Molar acetic acid and crystaline sodium acetate.

Here is a problem you may have to actually use one day. In biochemistry some enzymes need to be at a particular pH to work at maximum. You would choose a weak acid with a pkA close to the pH you need. (The pkA of acetic acid is 4.76.) The osmolarity (the total molar amount of dissolved materials) may be specified. (It is here. The total of acetic acid and sodium acetate should be at 0.15 Molar.)

It is most convenient to use the Henderson - Hasselbach equation for this, as it has a term that can be the ratio of the two materials. The form of the H-H equation does not matter, but the concentration of the conjugate ion will have to be greater than the concentration of the acid because the pH is greater than the pkA of the weak acid.

What we get from the H-H equation is the ratio of the two constituents. We can use that ratio as one of the equations in a two - equation - two - unknown setup to substitute one into the other and calculate the concentration of acetic acid, [HA], and the concentration of sodium acetate, [A_].

But we still have not answered the question, "Explain how to make 5 L of pH 5, 0.15 M acetic acid-sodium acetate buffer." We have a 1.00 Molar solution of acetic acid and crystals of (solid) sodium acetate. The way we have to measure the acetic acid is by measuring the volume of the more concentrated solution. The way to measure the sodium acetate is to weigh it. We would need (54.7885 x 5 = 273.9425) ml of acetic acid and (82.04 x 0.0952115 x 5 = 39.055757) grams of sodium acetate.

The real answer is that you need to weigh 39.1 g of sodium acetate, measure 274 ml of the 1.00 Molar acetic acid and put them into a 5 liter volumetric flask with enough water to dissolve the sodium acetate. Then fill the volumetric flast to the line with distilled water and mix the solution.

Back to Acid and Base math problem #20.


FOLLOWING ARE TITRATION PROBLEMS. ASSUME THAT THE pH INDICATOR WILL BE THE RIGHT ONE TO BALANCE THE AMOUNT OF ACID AND BASE. SOME OF THE TITRATIONS ARE NOT ACID -  BASE TITRATIONS. AGAIN ASSUME THERE IS AN INDICATOR THAT WILL TELL WHEN MOLAR AMOUNTS ARE MATCHED.

21. 23.45 mL of 0.275 M sodium hydroxide was used to titrate against mL of acetic acid. What was the concentration in M of acetic acid?

22. 17.05 mL of 0.247 M barium hydroxide was used to titrate against 10 mL of nitric acid. What was the concentration in M of nitric acid?

23. 35.79 mL of 0.275 M sodium hydroxide was used to titrate against 15 mL of sulfuric acid. What was the concentration in M of sulfuric acid?

24. 24.92 mL of 0.00199 M silver nitrate was used to titrate against 5 mL of sodium chloride solution. What was the concentration of NaCl?

ANSWERS TO PROBLEMS

21. 1.29 M <22. 0.842 M> 23. 0.328 M 24. 9.92 E-3 M

 

 


 

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