In a **Strong Acid-Strong Base Titration** we do not use an ICF Chart or ICE Chart to determine the pH of the solution.

**Concept:** Strong Acid-Strong Base Titration

**Example:** Calculate the pH of the solution resulting from the titration of 75.0 mL of 0.100 M HBrO_{4} with 55.0 mL of 0.100 M NaNH_{2}.

Let's take a look here where it says, 'Calculate the pH of the solution resulting from the titration of 75.0 mL of 0.100 M HBrO4 with 55 mL of 0.100 M NaNH2. Remember, this right here is a strong acid. This right here is a strong base. What we're going to say here is because both are strong, we don't have to use an ICE, we don’t have to use an ICF.

What do we do instead? We're going to say step one is find the moles. Remember, how do we find moles? Remember, moles equals molarity times liters. Divide the mL by a thousand, multiply them times their lit on their molarities.

Now, remember for the base, for the base you're always supposed to check how many OH minuses, NH2 minuses, H minuses and O2 minuses you have to get the correct concentration. Here we only have one NH2, so we don't have to worry about multiplying the concentration by anything, we can keep it as it is.

When we do this, we're going to say 0.075 L times 0.100 M HBrO4 equals 0.0075 M of HBrO4. Then we do the same thing with the other one, so we have that. Now that we found their moles, we're going to subtract the large one by the smaller one. What you're going to say is now you have this much left of your strong acid. That's all you're going to say, that's how much you have left of your strong acid.

Now step two, you're going to find molarity of whatever you have left. How do we find molarity? Well, we're going to take the moles that we just found and divide them by the total volume. Those two liters added up together will give you my total volume.

We're going to say 0.0020 moles of HBrO4 divide by the total volume we use, 0.0075 L plus 0.055 L. When we divide those together, that's going to give me, 0.0154 M HBrO4. Remember, since this is a strong acid, to find pH we don't need to do an ICE Chart. Since this is a strong acid, just take the negative log. Negative log of a strong acid will give me my pH.

So we plug in the concentration here and at the end, we scroll up some, at the end, what is that going to give me, my pH will be 1.81. So it's as simple as that. If both are strong, this is a great situation, first find the moles of both.

Remember look at the correct molarity of the strong base. If you have to multiply it by two because there's two OH or there's two NH2 or there's two Hs, H minuses, then you do that before you change everybody into moles.

Once you find moles, you subtract the larger moles from the smaller moles, whoever is left behind, whether it be a strong acid or a strong base, you immediately change it into molarity. If you have the molarity of a strong acid, take the negative log to find pH.

If you have the molarity of a base, so strong base then take the negative log of that, but that will give us pOH. So remember strong base if you take the negative log, it gives us pOH. If you have pOH, you're going to take that answer that you find and subtract it from 14 and that will give you pH. It's as simple as that.

Just remember the steps. Step one—find moles, step two—find molarity, step three—I guess you just take the negative log or whatever your answer is to find neither, pH or pOH. That's all you really have to do, there's no need of any type of chart.

Now that we've done this one, I want you guys to attempt to do the next one on your own as practice. So just take the same method that we used above in order to find it. Make sure that you get the correct concentrations of everyone before change everyone into moles. As long as you can remember that, you'll be able to solve this question. Good luck, guys.

If after the titration there is an excess of strong acid left then we can use its concentration to find the pH.

If after the titration there is an excess of strong base then we can use its concentration to find pOH initially.

**Problem:** Calculate the pH of the solution resulting from the mixing of 175.0 mL of 0.250 M HNO_{3} with 75.0 mL of 0.200M Ba(OH)_{2}.

If there is an equal amount of moles for both the strong acid and strong base then the pH of the solution will be 7.

What volume of 0.236 M H_{2}SO_{4} is needed to react with 53.5 mL of 0.104 M NaOH? The equation is

H_{2}SO_{4} aq) + 2NaOH (aq) → Na_{2}SO_{4} (aq) + 2H_{2}O (l)

Watch Solution

A patient is suspected of having low stomach acid, a condition known as hypochloridia. To determine whether the patient has this condition, her doctors take a 19.00 mL sample of her gastric juices and titrate the sample with 0.000452 M KOH. The gastric juice sample required 11.3 mL of the KOH titrant to neutralize it.

Calculate the pH of the gastric juice sample. Assume the sample contained no ingested food or drink which might otherwise interfere with the titration.

For the patient to be suffering from hypochloridia, the pH of the gastric juices from the stomach must be greater than pH 4. Does the patient have hypochloridia?

(a) No

(b) Yes

(c) Unable to determine

Watch Solution

What is the molarity of a sulfuric acid solution if it requires 62.55 mL of 0.09918 M NaOH to completely neutralize 25.00 mL of the acid?

H_{2}SO_{4} (aq) + 2NaOH (aq) → Na_{2}SO_{4} (aq) + 2H_{2}O (l)

Give your answer to 4 significant figures and DO NOT include the units.

Watch Solution

A 25.0 mL sample of 0.723 M HCIO_{4} is titrated with a 0.27 M KOH solution. The H_{3}O^{+} concentration after the addition of 80.0 mL of KOH is _____ M.

Watch Solution

Calculate the pH of the resulting solution if 20.0 mL of 0.200 M HCl(aq) is added to:

a. 30.0 mL of 0.200 M NaOH(aq)

b. 10.0 mL of 0.300 M NaOH(aq)

Watch Solution

Calculate the pH of the resulting solution if 20.0 mL of 0.200 M HCl(aq) is added to:

a. 30.0 mL of 0.200 M NaOH(aq)

b. 10.0 mL of 0.300 M NaOH(aq)

Watch Solution

100. mL of 0.200 *M* HCl is titrated with 0.250 *M* NaOH.

i) What is the pH of the solution after 50.0 mL of base has been added? Express the pH numerically.

ii) What is the pH of the solution at the equivalence point? Express the pH numerically.

Watch Solution

Complete and balance each of the following equations for acid-base reactions. Express your answer as a chemical equation. Identify all of the phases in your answer.

a) HBr(*a**q*) + LiOH(*a**q*) →

b) H_{2}SO_{4}(*a**q*) + Ba(OH)_{2}(*a**q*) →

Watch Solution

Consider the titration of 100.0 mL of 0.100 M NaOH with 1.00 M HBr. Find the pH at the following volumes of acid added: 0, 5,10,12 mL

Watch Solution

A 25.00 mL sample of 0.320 M NaOH analyte was titrated with 0.750 M HI at 25 °C.

a) Calculate the initial pH before any titrant was added.

b) Calculate the pH of the solution after 5.00 mL of the titrant was added.

Watch Solution

Determine the pH of a solution created by mixing 95.0 mL of 0.200 M nitric acid, HNO_{3}, with 320.0 mL of 0.078 M potassium hydroxide, KOH.

A. 2.22

B. 12.16

C. 0.91

D. 11.78

E. 1.84

Watch Solution

A student is titrating 38.0 mL of 0.0358 M solution of HBr with a 0.0289 M KOH solution. Calculate the pH of the solution after the addition of 43.19 mL of the KOH solution.

A. 2.53

B. 3.57

C. 12.46

D. 2.86

E. 1.78

Watch Solution

A 10. mL sample of 0.20 M hydrochloric acid solution is required to neutralize 20. mL of barium hydroxide, Ba(OH)_{2}

a. What is the molarity of the barium hydroxide solution?

b. What is the molarity of the salt that forms?

Watch Solution

A 10. mL sample of 0.20 M chloric acid solution is required to neutralize 20. mL of sodium hydroxide solution, NaOH.

a. What is the molarity of the sodium hydroxide solution?

b. What is the molarity of the salt that forms?

Watch Solution

What is the pH of a solution made by mixing 40.0 mL of 0.100 M HCl with 25.00 mL of 0.100 M KOH? Assume that the volumes of the solutions are additive.

A) 0.64

B) 1.64

C) 12.36

D) 13.36

E) 10.00

Watch Solution

A flask contains 50.0 mL of a 0.000968 M solution of Hydrochloric acid, HCl. The acid is to be titrated against 0.0275 M Sodium hydroxide, NaOH. What is the pH of the acid in the flask before the addition of any NaOH?

a. 1.32

b. 1.71

c. 3.01

d. 4.32

e. 1.56

Watch Solution

Calculate the pH of a solution resulting from the titration of 113.00 mL of 0.0442 M HCIO_{3} with 82.65 mL of 0.0624 M LiOH.

A. 10.92

B. 3.08

c. 10.21

D. 10.54

E. 7.00

Watch Solution

Calculate the pH of the solution resulting from the addition of 32.0 mL of 0.220 M HClO_{4} to 76.0 mL of 0.170 M NaOH.

1. 12.89

2. 7.00

3. 1.26

4. 1.11

5. 13.23

6. 12.74

7. 0.77

Watch Solution

A 100.0 mL sample of 0.18 M HBrO _{3} is titrated with 0.27 M LiH. Determine the pH of the solution after the addition of 30.0 mL of LiH.

A) 0.86

B) 1.21

C) 2.00

D) 1.12

E) 2.86

Watch Solution

Initially 7.1 g of HCl (MM = 36.46) and 9.5 g of NaOH (MM = 40.00) is added to water making a solution that has a volume of 12.85 L. What is the pH of the solution?

A. 1.82

B. 2.48

C. 7.00

D. 11.52

E. 12.27

Watch Solution