Calculating pH of Strong Bases

Learning Objectives (OCR Spec 5.1.3)

By the end of this lesson you should be able to:

• Define a strong base and give common examples
• Use Kw to find [H+] from [OH-] in a strong base solution
• Calculate the pH of a strong monobasic base (e.g. NaOH, KOH)
• Calculate the pH of a strong dibasic base (e.g. Ba(OH)2)
• Perform dilution and mixing calculations for strong bases

What is a Strong Base?

A strong base is one that is essentially fully ionised in aqueous solution, producing one mole of OH- per mole of base (per hydroxide group). The most important examples at A-Level are the Group 1 hydroxides and the heavier Group 2 hydroxides:

Base	Formula	Ionisation	OH- per formula unit
Sodium hydroxide	NaOH	Na+ + OH-	1
Potassium hydroxide	KOH	K+ + OH-	1
Lithium hydroxide	LiOH	Li+ + OH-	1
Calcium hydroxide	Ca(OH)2	Ca^2+ + 2OH-	2 (but limited solubility)
Barium hydroxide	Ba(OH)2	Ba^2+ + 2OH-	2

Group 1 hydroxides are very soluble strong bases. Ca(OH)2 is only slightly soluble (limewater) but what does dissolve is fully ionised. NH3(aq), by contrast, is a weak base - it does not fully ionise and needs its own Kb.

Strategy: Strong Base -> [OH-] -> [H+] -> pH

To calculate the pH of a strong base solution, we need [H+]. But the base gives us [OH-] directly. The bridge is Kw:

Kw = [H+][OH-] = 1.00 x 10^-14 mol^2 dm^-6 (at 298 K)

Rearranging: [H+] = Kw / [OH-]

Once we have [H+], pH = -log10[H+].

Worked Example 1: 0.100 mol dm-3 NaOH

Step 1: Find [OH-]. NaOH is fully ionised: 1 mol NaOH -> 1 mol OH-. So [OH-] = 0.100 mol dm-3.

Step 2: Find [H+]. [H+] = Kw / [OH-] = (1.00 x 10^-14) / 0.100 = 1.00 x 10^-13 mol dm-3

Step 3: Find pH. pH = -log10(1.00 x 10^-13) = 13.00

Worked Example 2: 0.0500 mol dm-3 KOH

[OH-] = 0.0500 mol dm-3 [H+] = (1.00 x 10^-14) / 0.0500 = 2.00 x 10^-13 mol dm-3 pH = -log10(2.00 x 10^-13) = 12.70 (2 dp)

Worked Example 3: 5.00 x 10^-3 mol dm-3 NaOH

[OH-] = 5.00 x 10^-3 [H+] = (1.00 x 10^-14) / (5.00 x 10^-3) = 2.00 x 10^-12 mol dm-3 pH = -log10(2.00 x 10^-12) = 11.70 (2 dp)

Worked Example 4: Strong Dibasic Base - Ba(OH)2

Calculate the pH of 0.0250 mol dm-3 Ba(OH)2. Assume complete dissociation.

Ba(OH)2 -> Ba^2+ + 2OH-

[OH-] = 2 x 0.0250 = 0.0500 mol dm-3 (one mole of Ba(OH)2 gives two moles of OH-) [H+] = (1.00 x 10^-14) / 0.0500 = 2.00 x 10^-13 mol dm-3 pH = -log10(2.00 x 10^-13) = 12.70 (2 dp)

Note that 0.0250 mol dm-3 Ba(OH)2 has the same pH as 0.0500 mol dm-3 KOH because each Ba(OH)2 delivers two OH-.

Worked Example 5: Dilution of a Strong Base

25.0 cm3 of 0.500 mol dm-3 NaOH is diluted to 250.0 cm3. Calculate the pH of the diluted solution.

c2 = (0.500 x 25.0) / 250.0 = 0.0500 mol dm-3 [OH-] = 0.0500 mol dm-3 [H+] = (1.00 x 10^-14) / 0.0500 = 2.00 x 10^-13 mol dm-3 pH = 12.70 (2 dp)

A 10x dilution drops pH by 1, just as it raises a strong acid pH by 1. The pH scale is symmetric around pH 7 in this sense.

Worked Example 6: Mixing Two Strong Bases

25.0 cm3 of 0.200 mol dm-3 NaOH is mixed with 75.0 cm3 of 0.0500 mol dm-3 Ba(OH)2. Find the pH.

Moles OH- from NaOH = 0.200 x 0.0250 = 5.00 x 10^-3 mol Moles OH- from Ba(OH)2 = 2 x 0.0500 x 0.0750 = 7.50 x 10^-3 mol (factor of 2!) Total moles OH- = 1.25 x 10^-2 mol Total volume = 0.100 dm3 [OH-] = 1.25 x 10^-2 / 0.100 = 0.125 mol dm-3 [H+] = (1.00 x 10^-14) / 0.125 = 8.00 x 10^-14 mol dm-3 pH = -log10(8.00 x 10^-14) = 13.10 (2 dp)

Mixing Strong Acid and Strong Base (Excess)

When a strong acid and a strong base are mixed, they neutralise mole-for-mole:

H+(aq) + OH-(aq) -> H2O(l)

If one is in excess, the pH is determined by the excess species.

Worked Example 7

50.0 cm3 of 0.100 mol dm-3 HCl is added to 40.0 cm3 of 0.150 mol dm-3 NaOH. Find the pH of the mixture.

Moles H+ = 0.100 x 0.0500 = 5.00 x 10^-3 mol Moles OH- = 0.150 x 0.0400 = 6.00 x 10^-3 mol