How do you determine the final pH of an acid/ base mixture?

 

Determine the pH of the following acid/ base solutions

No. Volume Concentration Acid + Volume Concentration Base

Answers
pH

Eg. 500 mL 0.01 M Sulfuric acid + 1.0 L 0.01 mol L-1 Aluminum hydroxide  12.1

1

50 mL 0.050 M Nitric acid + 50 mL 0.050 M Barium hydroxide  12.4

2

100 mL 0.20 mol L-1 Sulfuric acid + 100 mL 0.20 M Sodium hydroxide  1.0

3

500 mL 2.0 x 10-4 M Sulfuric acid + 500 mL 1.25 x 10-4 M Aluminum hydroxide  4.9

4

1.0 L 0.20 M Hydrochloric acid + 1.0 L 0.20 mol L-1  Sodium hydroxide  7.0

5

200 mL 0.01 M Hydrochloric acid + 100 mL 0.01 mol L-1 Potassium hydroxide  2.5

6

2.0 L 5.0 x 10-5 M Phosphoric acid + 1.0 L 0.00020 M Magnesium hydroxide  9.5

7

50 mL 0.0004 mol L-1 Nitric acid  acid + 50 mL 0.0025 M Calcium hydroxide  11.4

8

200 mL 0.0075 M Hydrochloric acid + 100 mL  0.01 mol L-1 Sodium hydroxide  2.8

9

100 mL 0.010 M Sulfuric acid + 50.0 mL 0.0125 M Barium hydroxide  2.3

10

100 mL 0.010 M Phosphoric acid + 100 mL 0.010 M Calcium hydroxide  2.3

 

Example.

Determine the final pH after adding 500mL of 0.01 M sulfuric acid to 1.0 L of 0.01 M aluminum hydroxide.

Note: The net ionic equation for this reaction is H+(aq) + OH-(aq)  H2O(l)
 

3H2SO4(aq) + 2Al(OH)3(aq) → Al2(SO4)3(aq) + 6H2O(l)
6H+(aq) +  3SO42-(aq) + 2Al3+(aq) + 6OH-(aq) → 2Al3+(aq) + 3SO42-(aq) + 6H2O(l)
6H+(aq) + 6OH-(aq) → 6H2O(l)
H+(aq) + OH-(aq) H2O(l)

 

PROCEDURE

1. The NET ionic equation for an acid/ base reactions is

H+(aq) + OH-(aq) → H2O(l)

 

2. Calculate the number of moles of H+ and OH- ions in each solution respectively.
 

(Don’t forget to dissociate acids and bases)
 

H2SO4
Al(OH)3

H2SO4 (aq) → 2H+(aq) + SO42-(aq)
0.01 M, 500 mL
n(H+) = 2 x c.V
n(H+) = 2 x 0.01 x 500/1000 mol
n(H+) = 0.01 mol
 

Al(OH)3(aq) → Al3+(aq) + 3OH-(aq)
0.01 M, 1.0 L
n(OH-) = 3 x c.V
n(OH-) = 3 x 0.01 x 1.0
n(OH-) = 0.03 mol
 

 

3. Determine the number of moles of excess H+(aq) or OH-(aq) ions.

If there is no excess the solution is neutral.

n(OH-) in excess = 0.03 mol - 0.01 mol

n(OH-) in excess = 0.02 mol
 

4. Calculate the final concentration of H+(aq) or OH-(aq) ions.

(Don’t forget to combine the volumes of both solutions)

V(Final) = V(H2SO4) + V(Al(OH)3)
V(Final) = 500 mL + 1.0 L
V(Final) = 1.5 L

c(OH-) = n(OH-)/V(Final)

c(OH-) = 0.02 mol/ 1.5 L

c(OH-) = 0.01333' mol.L-1

 

5. Calculate the pH of the final solution

(Formulae: pH = -log10[H+], pOH = -log10[OH-] & pH + pOH=14)

pOH = -log10[OH-]
pOH = -log10[0.01333']
pOH = 1.875

Since pH + pOH=14
pH = 14 - 1.875
pH = 12.1

 

Answer: pH =12.1

 

Student Common Errors:

A. Failure to dissociate in step 2.

B. Failure to combine the volumes in step 4.

 

Molar Mass
 

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