### What pH results when some strong acid and strong base solutions are mixed?Ten Examples

Comment before starting: in the examples below, you will find 1:1 molar ratios used in the first four examples and 2:1 molar ratios used in the next three.

Here is a possibility: the teacher may only show examples using 1:1 ratios (since they are, by far, the most common) and then ask a 2:1 ratio on the test.

Or, the teacher may include some 2:1 ratios and then have a 3:1 or 3:2 ratio on the test. I have included two 3:1 ratios and a 3:2 ratio for the last three examples.

Since this set of examples and problems concerns ONLY strong acids and bases, you will see HCl and NaOH over and over and over. Sometimes, HBr or KOH are used, but there is no difference in the solution technique when that happens.

H2SO4 and Ba(OH)2 are the most common acid and base used in 2:1 ratios.

Since the reactants are all strong electrolytes, everything is dissociated 100%. Sulfuric acid is not strong in its second hydrogen, but that is commonly ignored. Often, the problems are rigged so that H2SO4 is not the excess reagent.

These problems can be solved using millimoles and milliliters rather than moles and liters. Some of the problems below are solved using millimoles and some using moles.

Example #1: Calculate the pH of the solution that results when 20.0 mL of 0.600 M HCl is reacted with 20.0 mL of 0.600 M NaOH solution.

Solution using moles:

1) Write the balanced, chemical equation:

HCl + NaOH ---> NaCl + H2O

2) From the balanced equation, determine the molar ratio of the two reactants:

HCl and NaOH are the reactants

the molar ratio is 1:1

3) For each of the two reactants, compute the moles present before reaction:

HCl ---> (0.600 mol/L) (0.0200 L) = 0.0120 mol
NaOH ---> (0.600 mol/L) (0.0200 L) = 0.0120 mol

4) Compute the number of moles of each reactant after reaction:

The reactants react in a 1:1 molar ratio. We note that 0.0120 to 0.0120 is a 1:1 molar ratio. The two reactants mutually neutralize each other with no left-over HCl and no left-over NaOH.

5) The remaining solution is composed of one solute, that being sodium chloride. Since NaCl is the salt of a strong acid and a strong base, it exerts no influence on the pH of the solution. No calculation is required to determine that the pH of the remaining solution is 7.000.

Example #2: Calculate the pH of the solution that results when 20.0 mL of 0.600 M HCl is reacted with 25.0 mL of 0.600 M NaOH solution.

Solution using millimoles:

1) For each of the two reactants, compute the millimoles present before reaction:

HCl ---> (0.600 mmol/mL) (20.0 mL) = 12.0 mmol
NaOH ---> (0.600 mmol/mL) (25.0 mL) = 15.0 mmol

Note that molarity expressed using millimoles is millimole/milliliter, which is exactly equivalent to mol/L

2) HCl and NaOH react in a 1:1 molar ratio, which means we can simply subtract the smaller amount from the larger:

15.0 − 12.0 = 3.0 mmole of NaOH remains

3) Compute the molarity of the NaOH in the combined volume:

3.0 mmol / 45.0 mL = 0.066667 M

4) Compute the pOH:

pOH = −log 0.066667

pOH = 1.176

5) Compute the pH:

pH + pOH = 14

pH = 14 − 1.176 = 12.824

Example #3: Calculate the pH of the solution that results when 25.0 mL of 0.600 M HCl is reacted with 20.0 mL of 0.600 M NaOH solution.

Solution using millimoles:

1) For each of the two reactants, compute the millimoles present before reaction:

HCl ---> (0.600 mmol/mL) (25.0 mL) = 15.0 mmol
NaOH ---> (0.600 mmol/mL) (20.0 mL) = 12.0 mmol

2) HCl and NaOH react in a 1:1 molar ratio, which means we can simply subtract the smaller amount from the larger:

15.0 − 12.0 = 3.0 mmole of HCl remains

3) Calculate the molarity:

3.0 mmole / 45.0 mL = 0.066667 M

4) Calculate the pH:

pH = −log 0.066667

pH = 1.176

Example #4: A 50.0 mL sample of 0.50 M HCl is titrated with 0.50 M NaOH. What is the pH of the solution after 28.0 mL of NaOH have been added to the acid?

Solution:

1) Calculate moles of HCl and NaOH:

moles HCl ⇒ (0.50 mol/L) (0.050 L) = 0.025 mol
moles NaOH ⇒ (0.50 mol/L) (0.028 L) = 0.014 mol

2) Calculate moles HCl remaining:

Since HCl and NaOH react in a 1:1 molar ratio:

0.025 mol − 0.014 mol = 0.011 mol HCl remaining

3) Calculate [HCl] of new solution:

0.011 mol / 0.078 L = 0.141 M

Note volume of 78 mL, derived from 50 + 28.

4) Calculate pH:

pH = −log [H+]

Since HCl dissociates 100%:

pH = −log 0.141 = 0.85

Example #5: Consider the titration of 80.0 mL of 0.100 M Ba(OH)2 by 0.400 M HCl. What is the pH of the solution (a) before adding any acid and (b) after adding 20.0 mL of HCl?

Solution:

1) Determine [OH¯], then pH:

Ba(OH)2 ---> Ba2+ + 2OH¯

Ba(OH)2 is a strong base, so it dissociates 100%. Remember also, that two hydroxides are produced for every Ba(OH)2.

[OH¯] = 0.200 M

pOH = −log 0.200 = 0.699

pH = 14.000 − pOH = 14.000 − 0.699 = 13.301 (the answer to part a)

2) Determine moles of OH¯ remaining after 20.0 mL of HCl is added:

moles H+ ⇒ (0.400 mol/L) (0.0200 L) = 0.00800 mol

(Remember, HCl dissociates 100%.)

moles [OH¯] ⇒ (0.200 mol/L) (0.0800 L) = 0.0160 mol

(Note use of 0.200 M for [OH¯]; the [Ba(OH)2] is 0.100 M.)

remaining OH¯ ⇒ 0.0160 mol − 0.00800 mol = 0.00800 mol

3) Determine [OH¯], then pH:

[OH¯] = 0.00800 mol / 0.100 L = 0.080 M

(Note the use of 100 mL, which comes from 80 + 20. There is an asumption that the volumes are additive. In context − both solutions are aqueous − this is a pretty safe assumpton to make.)

pOH = −log 0.080 = 1.097

pH = 14.000 − 1.097 = 12.903 (the answer to part b)

Example #6: Hydrochloric acid (50.0 mL of 0.446 M) is added to 250.0 mL of 0.0560 M Ba(OH)2 solution. What is the concentration of the excess H+ or OH¯ ions left in this solution?

Solution (not the usual ChemTeam approach):

HCl and Ba(OH)2 are both strong electrolytes, so they each ionize 100% in solution.

moles H+ from HCl ---> (0.446 mol) (0.0500 L) = 0.0223 mol
moles OH¯ from Ba(OH)2 ---> (0.0560 mol/L) (0.2500 L) (2) = 0.0280 mol

The 2 is present in the second calculation because Ba(OH)2 produces 2 hydroxide ions for every one Ba(OH)2 formula unit that dissolves and ionizes. Since we now have the total moles of H+ and OH¯, we will react them in a 1:1 molar ratio (not a 2:1 ratio), based on the following reaction:

H+ + OH¯ ---> H2O

Since the OH¯ is in excess, we do this:

0.0280 − 0.0223 = 0.0057 mol

to determine the OH¯ left over.

For the concentration of the OH¯, we use the total volume of 300.0 mL:

0.0057 mol / 0.3000 L = 0.019 M

Although the pH is not asked for, it can easily be calculated:

pOH = −log 0.019 = 1.72

pH = 14 − 1.72 = 12.28

Example #7: A 100.0 mL sample of 0.10 mol L¯1 Ca(OH)2 is titrated with 0.10 mol L¯1 HBr. Determine the pH of the solution after the addition of 400.0 mL HBr.

Solution

1) Important starting information:

Ca(OH)2 + 2HBr ---> CaBr2 + 2H2O

Mole ratio in reaction: Ca(OH)2 to HBr = 1:2

2) Moles of each reactant present:

Ca(OH)2 ---> (0.10 mol/L) (0.1000 L) = 0.010 mol
HBr ---> (0.10 mol/L) (0.4000 L) = 0.040 mol

3) Determine limiting reagent:

Ca(OH)2 ---> 0.010 mol / 1 = 0.01
HBr ---> 0.040 mol / 2 = 0.02

Ca(OH)2 is the limiting reagent, therefore HBr is in excess.

4) Determine amount of HBr remaining:

For every one mole of Ca(OH)2 that reacts, two moles of HBr are consumed. Therefore:

(0.010 mol) (2) = 0.020 mol of HBr consumed

0.040 mol − 0.020 mol = 0.020 mol HBr remaining

5) Calculate pH of solution (HBr is a strong acid):

0.020 mol / 0.500 L = 0.040 M

pH = −log 0.040 = 1.40

6) You might want to try this problem using millimoles. Here's the starting calculation:

Ca(OH)2 ---> (0.10 mmol/mL) (100.0 mL) = 10 mmol
HBr ---> (0.10 mmol/mL) (400.0 L) = 40 mmol

Example #8: Let us suppose 22.30 mL of a solution of 0.0450 M HCl is reacted with 38.70 mL of 0.00818 M M(OH)3 (a made-up formula that ionizes 100% and is soluble). What is the final pH of the solution? (I know that aluminum hydroxide exists. However, I want a soluble base that is tribasic, so I made one up.)

Solution:

1) Determine total moles of hydrogen ion and hydroxide ion:

HCl ---> (0.0450 mmol/mL) (22.30 mL) (1) = 1.0035 mmol
M(OH)3 ---> 0.00818 mmol/mL) (38.70 mL) (3) = 0.949698 mol

The (1) is due to the fact that one H+ is formed for every one HCl that dissolves and the (3) is due to the fact that three hydroxide are formed for every one M(OH)3 that dissolves.

2) H+ and OH¯ always react in a 1:1 molar ratio. Determine how much of the excess remains after reaction:

1.0035 mmol − 0.949698 mol = 0.053802 mmol of H+

3) Determine the new [H+]:

0.053802 mmol / 61.0 mL = 0.000882 M

4) Calculate the pH:

pH = −log 0.000882 = 3.054

Example #9: Let us suppose we have a solution of a triprotic acid (call its formula H3A) that is strong in all three of its hydrogens. We mix 45.0 mL of 0.0675 M H3A solution with 28.0 mL of 0.0985 M NaOH. What is the pH of the resulting solution?

Solution using millimoles:

1) Write the chemical reaction:

H3A + 3NaOH ---> Na3A + 3H2O

The 1 to 3 molar ratio between acid and base will come into play below.

2) Calculate moles of H3A and Ba(OH)2:

H3A ⇒ (0.0675 mmol/mL) (45.0 mL) = 3.0375 mmol
NaOH ⇒ (0.0985 mol/mmL) (28.0 mL) = 2.758 mmol

3) Determine moles of H+ and OH¯ in solution before reaction (here is where the 3:1 molar ratio is used):

H+ ⇒ (3.0375 mmol) (3) = 9.1125 mmol
NaOH ⇒ (2.758 mmol) (1) = 2.758 mmol

4) The H+ and the OH¯ in solution react in a 1:1 molar ratio. How much of the excess reagent remains?

9.1125 mmol − 2.758 mmol = 6.3545 mmol of H+

5) Calculate the pH after reaction has ceased:

45.0 mL + 28.0 mL = 73.0 mL

6.3545 mmol / 73.0 mL = 0.087048 M (keep some extra digits)

pH = −log 0.087048 = 1.060

Example #10: Let us suppose we have a solution of a triprotic acid (call its formula H3A) that is strong in all three of its hydrogens. We mix 35.0 mL of 0.0875 M acid solution with 22.0 mL of 0.0585 M Ba(OH)2. What is the pH of the resulting solution?

Solution #1:

1) Write the chemical reaction:

2H3A + 3Ba(OH)2 ---> Ba3A2 + 6H2O

The 2 to 3 molar ratio between acid and base will come into play below.

2) Calculate moles of H3A and Ba(OH)2:

H3A ⇒ (0.0875 mol/L) (0.0350 L) = 0.0030625 mol
Ba(OH)2 ⇒ (0.0585 mol/L) (0.0220 L) = 0.001287 mol

3) Determine the limiting reagent:

H3A ⇒ 0.0030625 / 2 = 0.00153125
Ba(OH)2 ⇒ 0.001287 / 3 = 0.000429

The Ba(OH)2 is limiting.

4) Determine how much H3A is used up:

 2 x ––––––– = ––––––– 3 0.001287

x = 0.000858 mol

5) Determine how much H3A remains:

0.0030625 − 0.000858 = 0.0022045 mol

6) Determine the new molarity of the H3A:

0.0022045 mol / 0.0570 L = 0.038675 M <--- I'll keep two extra digits for a moment

7) Since H3A ionizes 100% in all three hydrogens, it produces a [H+] equal to 0.116025 M. Determine the pH:

pH = −log 0.116025 = 0.935

Solution #2:

1) Determine the total moles of H+ and OH¯ delivered to the solution:

H3A ⇒ (0.0875 mol/L) (0.0350 L) (3) = 0.0091875 mol
Ba(OH)2 ⇒ (0.0585 mol/L) (0.0220 L) (2) = 0.002574 mol

This is NOT the total moles of H3A and Ba(OH)2. It is the total moles of H+ and OH¯ in the solution AFTER H3A and Ba(OH)2 have each ionized 100%.

The 2 and the 3 in the calculations take care of the 2:3 molar ratio that the molecular species react in.

2) H+ and OH¯ react in a 1:1 molar ratio:

H+ + OH¯ ---> H2O

Remember, it is the molecular species [H3A and Ba(OH)2] that react in a 2:3 ratio

3) In a 1:1 molar ratio, the smaller amount is the limiting reagent. Determine amount of excess reagent remaining after reaction ceases:

0.0091875 − 0.002574 = 0.0066135 mol of H+ remaining

4) Determine molarity of H+:

0.0066135 mol / 0.0570 L = 0.116026 M

5) Determine pH:

pH = −log 0.116026 = 0.935

Bonus Example #1: A 0.15 mole quantity of NaOH is dissolved in 1.50 L of water. In a separate container, 0.15 mole of HCl is present in 1.50 L of water. What is the pH of a solution made by mixing the two?

Solution:

What you have to do is determine how much NaOH or HCl is left over after the two solutions are mixed.

In this problem, the HCl and NaOH are present in the exact amounts have no HCl or NaOH left over when the reaction is complete. This is because the two react in a 1:1 molar ratio, as seen in the chemical equation:

HCl + NaOH ---> NaCl + H2O

There is 0.15 mole of NaOH available to react and there is 0.15 mole of HCl available to react. They are present in a 1:1 molar ratio and neither NaOH or HCl is left over at the end of the reaction.

The mixed solution in your problem has only NaCl dissolved. Since this is the salt of a strong acid and a strong base, there is no effect on the pH of the solution from the NaCl.

The pH of the mixed solution is 7.

Bonus Example #2: 8.00 g of Ca(OH)2 (Ksp = 5.5 x 10¯6) is mixed in to 40.0 mL of water to create a saturated solution. If 10.0 mL of 0.100 M HCl is added to this solution, what is the resulting pH?

Solution:

Comment: Not all of the 8 g dissolved. If all 8 g dissolved, we would not know if the solution was saturated. One of the hallmarks of a saturated solution is that not all the solid present dissolves.

1) We will start by calculating the molar solubility of Ca(OH)2.

Ksp = [Ca2+] [OH¯]2

5.5 x 10¯6 = (s) (2s)2

s = 0.0111199 mol/L

2) How many millimoles of Ca(OH)2 are in the 40.0 mL?

(0.0111199 mmol/mL) (40.0 mL) = 0.444796 mmol

3) How many mmoles of just hydroxide are in the 40.0 mL?

(0.444796 mmol) (2) = 0.889592 mmol

That 2 comes from the fact that there are two hydroxides in solution for every Ca(OH)2 that dissolved.

4) Determine millimoles of H+ in solution:

(0.100 mmol/mL) (10.0 mL) = 1.00 mmol

5) The reaction between hydrogen ion and hydroxide ion is this:

H+ + OH¯ ---> H2O

Note the 1:1 molar ratio between reactants.

6) The hydrogen ion is in excess. Determine the amount remaining after reaction ceases:

1.00 mmol − 0.889592 mmol = 0.110408 mmol

7) Determine the molarity of the H+:

0.110408 mmol / 50.0 mL = 0.00220816 M

8) Determine the pH:

pH = −log 0.00220816 = 2.656