What pH results when some strong acid and strong base solutions are mixed?
Problems #1-10

Problems 11-25

Ten Examples

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Problems 1 to 5 and 9 use 1:1 ratios. Problems 6, 8, and 10 are 2:1 ratios and problem 7 is a 3:2 ratio.


Problem #1: A 50.0 mL volume of 0.150 M HBr is titrated with 0.250 M KOH. Calculate the pH after the addition of 11.0 mL of KOH.

Solution:

1) Determine excess reagent:

HBr ---> (0.150 mol/L) (0.0500 L) = 0.00750 mol
KOH ---> (0.250 mol/L) (0.0110 L) = 0.00275 mol

HBr(aq) + KOH(aq) ---> KCl(aq) + H2O(ℓ)

HBr and KOH react in a 1 to 1 molar ratio. Since HBr is greater in amount, it is the excess reagent.

2) Determine amount of HBr remaining after reaction:

0.00750 mol − 0.00275 mol = 0.00475 mol

3) Determine molarity of remaining HBr and the pH:

0.00475 mol divided by 0.0610 L = 0.077869 M

pH = −log 0.077869 = 1.109


Problem #2: At 298 K, 25.0 cm3 of a solution of strong acid contained 0.00150 moles of hydrogen ions. Calculate the pH of the solution formed after the addition of 50.0 cm3 of 0.150 M NaOH to the original 25.0 cm3 of acid.

Solution:

1) We know how many moles of hydrogen ion are present before reaction. We need to determine the moles of hydroxide that are present before reaction:

moles = MV ---> (0.150 mol/L) (0.0500 L) = 0.00750 mol

I won't bother showing the cm3 to mL to L conversion.

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

H+ + OH¯ ---> H2O

3) Clearly, the hydroxide is in excess. Determine the left over hydroxide:

0.0075 mol − 0.0015 mol = 0.0060 mol
4) The next step is to determine the molarity of the hydroxide in the volume of the two solutions when mixed:
0.0060 mol / 0.0750 L = 0.080 M

5) Now, for the pOH:

pOH = −log 0.080 = 1.09691

6) And the pH:

14 − 1.09691 = 12.90309

Three sig figs is the best answer:

12.903


Problem #3: What is the pH in the titration of 90.0 mL of 0.100 M HCl after the addition of 85.0 mL 0.100 M NaOH?

Solution:

1) Determine excess reagent:

HCl ---> (0.100 mol/L) (0.0900 L) = 0.00900 mol
NaOH ---> (0.100 mol/L) (0.0850 L) = 0.00850 mol

HCl(aq) + NaOH(aq) ---> NaCl(aq) + H2O(ℓ)

HCl and NaOH react in a 1 to 1 molar ratio. Since HCl is greater in amount, it is the excess reagent.

2) Determine amount of HCl remaining after reaction:

0.00900 mol − 0.00850 mol = 0.00050 mol

3) Determine molarity of remaining HCl and the pH:

0.00050 mol / 0.175 L = 0.002857 M

pH = −log 0.002857 = 2.544

4) Comment: after learning the basics of this type of calculation, this approach can be seen to be viable (but only if the molarities are equal):

(0.100 mol/L) (5.0 mL) = (x) (175 mL)

x = 0.002857 M

and thence to the pH.

The 5.0 comes from 90.0 − 85.0

Also, you can make this technique work when the ratio is other than 1 to 1, but it's usually easier to determine the moles of each reactant and then compute the molarity of the excess reagent.


Problem #4: A 20.0 mL sample of 1.50 M HCl is titrated with 1.00 M NaOH. Calculate the pH after the following volumes of base have been added:

(a) 29.0 mL; (b) 30.0 mL; (c) 31.0 mL

First, I'll write the chemical equation (as if you don't already know it):

HCl(aq) + NaOH(aq) ---> NaCl(aq) + H2O(ℓ)

Second, let's do the HCl amount (in millimoles):

(1.50 mmol/mL) (20.0 mL) = 30.0 mmol

Solution to (a):

1) Determine mmol of NaOH:

(1.00 mmol/mL) (29.0 mL) = 29.0 mmol

2) Determine amount of HCl left over:

30.0 − 29.0 = 1.0 mmol

3) Determine molarity of the remaining HCl:

1.0 mmol / 49.0 mL = 0.020408 M

4) Determine pH:

pH = −log 0.020408 = 1.69

Solution to (b):

1) Determine mmol of NaOH:

(1.00 mmol/mL) (30.0 mL) = 30.0 mmol

2) Determine amount of HCl left over:

30.0 − 30.0 = 0.0 mmol

HCl and NaOH were in stoichiometrically equal amounts, There was no limiting reagent.

3) The solution is one of NaCl, the salt of a strong acid and a strong base. This type of salt forma neutral solution so, by definition, the pH is 7.00.

Solution to (c):

1) Determine mmol of NaOH:

(1.00 mmol/mL) (31.0 mL) = 31.0 mmol

2) Determine amount of NaOH left over:

31.0 − 30.0 = 1.0 mmol

3) Determine molarity of the remaining NaOH:

1.0 mmol / 51.0 mL = 0.019608 M

4) Determine pOH:

pOH = −log 0.019608 = 1.71

5) Determine pH:

14 − 1.71 = 12.29

Problem #5: Consider the titration of 20.0 mL of 0.100 M perchloric acid with 0.400 M LiOH. Calculate the pH after the following percentages of titrant has been added to the analyte.

(a) 0.00%   (b) 25.0%   (c) 50.0%   (d) 75.0%   (e) 100.0%   (f) 110.0%

Solution:

1) Information needed:

moles HClO4 ---> ( 0.100 mol/L) (0.0200 L) = 0.00200 mol

HClO4 + LiOH ---> LiClO4 + H2O

HClO4 and LiOH react in a one to one molar ratio.

2) Part (a):

Since HClO4 is a strong acid:

[H+] = 0.100 M

pH = −log 0.100 = 1.000

3) Part (b) (25% of the moles of perchloric acid):

moles LiOH required ---> (0.00200 mol) (0.25) = 0.00050 mol

volume LiOH required ---> 0.00050 mol / 0.400 mol/L = 0.00125 L

moles H+ in excess ---> 0.00200 mol − 0.00050 mol = 0.0015 mol

total volume ---> 0.00125 L + 0.0200 L = 0.02125 L

[H+] = 0.0015 mol / 0.02125 L = 0.070588 M

pH = −log 0.070588 = 1.151

4) Part (c) (50% of the moles of perchloric acid):

moles LiOH ---> (0.00200 mol) (0.50) = 0.00100 mol

volume LiOH ---> 0.00100 mol / 0.400 mol/L = 0.00250 L

total volume ---> 0.00250 L + 0.0200 L = 0.0225 L

moles H+ in excess ---> 0.00200 mol − 0.00100 mol = 0.00100 mol

[H+] = 0.00100 / 0.0225 = 0.044444 M

pH = 1.352

5) Part (d) (75% of the moles of perchloric acid):

the solution is left to the reader.

pH = 1.68

6) Part (e) (100% of the moles of perchloric acid):

moles LiOH = moles HClO4

The solution is of LiClO4 only.

Since LiClO4 is the salt of a strong acid and a strong base, the pH = 7.000

7) Part (f) (110% of the moles of perchloric acid):

moles LiOH ---> (0.00200 mol) (1.10) = 0.00220 mol

volume LiOH ---> 0.00220 mol / 0.400 mol/L = 0.00550 L

moles OH- in excess ---> 0.00220 − 0.00200 = 0.000200

total volume ---> 0.00550 L + 0.0200 L = 0.0255 L

[OH-] = 0.00200 mol / 0.0255 L = 0.00784314 M

pOH = −log 0.00784 = 2.106

pH = 11.894


Problem #6: A solution was made by mixing 250. mL of 0.30 M H2SO4 with 400. mL of 0.70 M NaOH. Calculate the pH after the reaction ceases.

Solution:

Comment: this is a limiting reagent problem. We need to figure out which substance is in excess and how much of it remains.

1) Moles of each reactant:

moles H2SO4 ---> (0.30 mol/L) (0.250 L) = 0.075 mol

moles NaOH ---> (0.70 mol/L) (0.400 L) = 0.28 mol

2) NaOH reacts with H2SO4 in a 2 to 1 ratio.

Therefore, 0.075 mol of H2SO4 consumes 0.15 mol of NaOH

3) Determine moles of NaOH that remain after H2SO4 is completely used up:

0.28 − 0.15 = 0.13 mol

4) Determine the NaOH concentration after reaction:

0.13 mol / 0.650 L = 0.20 M

5) Based on 100% ionization of the NaOH, the [OH¯] = 0.20 M. Determine the pH:

pOH = −log 0.20 = 0.70

pH = 14 − 0.70 = 13.30


Problem #7: How many mL of 0.400 M H2SO4 would require the same amount of base to reach full neutralization as would 25.0 mL of 0.600 M H3PO4?

Solution:

1) Phosphoric acid is not a strong acid, but let's act like it is. Determine the total amount of hydrogen ion in 25.0 mL of 0.600 M H3PO4:

(0.600 mmol/mL) (25.0 mL) (3) = 45.0 mmol

We can act like H3PO4 is a strong acid because we know all of the hydrogen ion in the H3PO4 will react with the base.

2) The question now is how much volume of the H2SO4 solution contains 45.0 mmol:

(0.400 mmol/mL) (x) (2) = 45.0

x = 56.25 mL <--- I won't bother to round it off.

3) Each amount of acid contains 45.0 mmol. 25.0 mL of 0.600 M phosphoric acid contains 45.0 mmol. 56.25 mL of 0.400 M sulfuric acid contains 45.0 mmol. Each requires the same amount of base because hydrogen ion and hydroxide ion react in a 1:1 molar ratio:
H+ + OH¯ ---> H2O

Problem #8: 50.0 mL of 0.209 M hydrochloric acid is added to 250.0 mL of 0.0555 M Ba(OH)2 solution. What is the concentration of the excess H+ or OH¯ ions left in this solution?

Solution:

1) The chemical equation:

2HCl + Ba(OH)2 ---> BaCl2 + 2H2O

2) Calculate moles of each reactant:

moles HCl ---> (0.209 mol/L) (0.050 L) = 0.01045 mol
moles Ba(OH)2 ---> (0.0555 mol/L) (0.250 L) = 0.013875 mol

3) Which is the limiting reagent?

HCl ---> 0.01045 / 2 = 0.005225
Ba(OH)2 ---> 0.013875 / 1 = 0.013875

HCl runs out first.

4) How much Ba(OH)2 is used up?

There is a 2:1 molar ratio between HCl and Ba(OH)2

0.01045 is to 2 as x is to 1

x = 0.005225 mol of Ba(OH)2 is used up

5) How much Ba(OH)2 remains?

0.013875 − 0.005225 = 0.00865 mol

6) What is the new molarity of the Ba(OH)2?

0.00865 mol / 0.300 L = 0.02883333 M

0.300 L is the total volume after the solutions are mixed. I didn't round the answer off because there is one more step.

7) You want the concentration of the hydroxide ions, not the concentration of the Ba(OH)2. There are two OH¯ ions per one formula unit of Ba(OH)2, so:

(0.02883333 M) (2) = 0.05766666 M

Three sig figs = 0.0577 M

8) The pH can be determined and it is 12.761. I will leave you, gentle reader, to do the math that results in that answer.


Problem #9: Consider a solution made of: (a) 230. mg KOH, (b) 8.71 mL of an 0.802 M HCl solution, (c) 3.00 g of a 3.97 %(w/w) solution of NaOH, and then diluted to a final volume of 618 mL What is the pH of this solution?

Solution:

1) Determine the moles of each solute:

(a) KOH:
230. mg = 0.230 g

0.230 g / 56.1049 g/mole = 0.0040995 mol

(b) HCl

(0.802 mol/L) (0.00871 L) = 0.00698542 mol

(c) NaOH

(3.00 g) (0.0397) = 0.1191 g (of the 3.00 g of solution is NaOH)

0.1191 g / 39.9969 g/mol = 0.00297773 mol

2) The HCl reacts with the KOH in a 1:1 molar ratio:

0.00698542 mol − 0.0040995 mol = 0.00288592 mol <--- HCl that remains

3) I note that there is slightly more NaOH than HCl. I know that NaOH and HCl react in a 1:1 molar ratio:

0.00297773 mol − 0.00288592 mol = 0.00009181 mol <---NaOH that remains

4) Determine the pOH, then the pH:

0.00009181 mol / 0.618 L = 0.00014856 M

pOH = −log 0.00014856 = 3.828

pH = 14 − 3.828 = 10.172

5) Since the KOH and the NOH both react with the HCl in a 1:1 molar ratio, I could have added the moles of KOH and NaOH, then subtracted the moles of HCl.


Problem #10: Calculate the pH of a solution made by mixing 50.0 mL of Ca(OH)2 with a mass concentration of 0.741 grams/liter with 50.0 mL of HCl with a concentration of 0.010 mol/liter.

Solution:

1) When HCl and Ca(OH)2 solutions react, we must first determine the number of moles of each substance first. Then, we can use the 2:1 molar ratio of the two reacting substances to determine how much is left over and thence to the pH.

moles Ca(OH)2 ---> (0.741 g/L / 74.0918 g/mol) (0.050 L) = 5.0 x 10¯4 mol
moles HCl ---> (0.010 mol/L) (0.0500 L) = 5.0 x 10¯4 mol

2) HCl is the limiting reagent:

Ca(OH)2 ---> 5.0 x 10¯4 / 1 = 5.0 x 10¯4
HCl ---> 5.0 x 10¯4 / 2 = 2.5 x 10¯4

3) Since two moles of HCl are required for every one mole of Ca(OH)2 neutralized, this results:

5.0 x 10¯4 mol of HCl will neutralize 2.5 x 10¯4 mol of Ca(OH)2.

That leaves 2.5 x 10¯4 mol of Ca(OH)2 in 100.0 mL of solution.

4) The molarity of the Ca(OH)2 is this:

2.5 x 10¯4 mol / 0.1000 L = 0.0025 M

5) When Ca(OH)2 ionizes in solution, it ionizes 100%. Which means there are two hydroxde ions produced for every one Ca(OH)2 dissolved. This means that the molarity of just the hydroxide ion is 0.0050 M. We now have the molarity we need.

pOH = −log 0.0050 = 2.30

pH = 14.00 − 2.30 = 11.70


Bonus Problem #1: A 50.00 mL of 0.172 M HNO3 was titrated with 0.206 M NaOH. Determine pH of the solution at the midway point during the titration.

Solution:

1) Determine the mmoles of nitric acid present before reaction:

(0.172 mmol/mL) (50.00 mL) = 8.60 mmol

2) 4.30 mmol of HNO3 reacted.

Leaving 4.30 mmol of nitric acid unreacted.

3) NaOH and HNO3 react in a 1:1 molar ratio.

This means 4.30 mmol of NaOH reacted.

4) What volume of 0.206 M NaOH will deliver 4.30 mmol?

(0.206 mL) (x) = 4.30 mmol

x = 20.8738 mL

5) Determine the new HNO3 molarity:

50.00 mL + 20.8738 mL = 70.8738 mL

4.30 mmol / 70.8738 mL = 0.06067122 M

6) Determine the pH:

pH = −log 0.06067122 = 1.217

Bonus Problem #2: A 1.00 liter solution contains 0.250 M NH3, 0.200 moles of NaOH, and 0.600 M NH4Br. What is the pH of the solution?

Comment: The trap in this problem is to see the strong base (NaOH) as overwhelming the others and then calculating the pH based on just the NaOH. That way ignores the fact that the NH4+ is an acid and reacts with the NaOH, but not the NH3.

Solution:

1) The NaOH reacts with the NH4+ in a 1:1 molar ratio. What are the concentrations of the NH4+ and the NH3 after reaction is complete?

The NaOH is the limiting reagent.

moles NH4+ present before reaction ---> (0.600 mol/L) (1.00 L) = 0.600 mol

moles NH4+ present after reaction ---> 0.600 mol − 0.200 mol = 0.400 mol

The new molarity of the ammonium ion is 0.400 M

moles NH3 present after reaction ---> 0.250 mol + 0.200 mol = 0.450 mol

The new molarity of the ammonia is 0.450 M.

2) Use the Henderson-Hasselbalch equation to determine the pH of the solution from the perspective of the ammonium ion.

This is the chemical reaction of interest:
NH4+ + H2O ⇌ NH3 + H3O+

The Henderson-Hasselbalch Equation:

pH = pKa + log [base / acid]

pH = 9.248 + log (0.450 / 0.400)

pH = 9.299

3) Use the Kb expression to determine the pH of the solution from the perspective of the ammonia.

This is the reaction of interest:
NH3 + H2O ⇌ NH4+ + OH¯

The Kb for ammonia is this:

  [NH4+] [OH¯]
Kb = –––––––––––
  [NH3]

  [0.400] [OH¯]
1.77 x 10¯5 = –––––––––––
  [0.450]

[OH¯] = 1.99125 x 10¯5 M (keep some extra digits)

pOH = 4.700874

pH = 9.299


Problems 11-25

Ten Examples

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