L4 - pH of Water at Different Temperatures

objectives

Understandings:

• For a conjugate acid base pair, Ka × Kb = Kw.

Applications and skills:

• Solution of problems involving [H+ (aq)], [OH–(aq)], pH, pOH, Ka, pKa, Kb and pKb.

Guidance:

• The value Kw depends on the temperature.

• Only examples involving the transfer of one proton will be assessed.

• Calculations of pH at temperatures other than 298 K can be assessed.

Kw and Temperature

H₂O ⇌ H⁺ + OH^- ΔH = +ve

From SL Chemistry we know that Kw = 1x10^-14 However this is only at room temperature.

We also know from our work on Equilibria Law that Temperature is the only variable that changes K.

Therefore it stands to reason that the value of Kw (and therefore the pH of water) can change for different temperatures:

• The relationship above shows us that the dissociation of water is endothermic
• This means that at higher temperatures the equilibrium position will move to the right
• This means that Kw will have a higher value and that water will have a lower pH

Worked Example

Calculate the pH of water at 40C when Kw = 2.09 x 10-14 mol2 dm-6 .

Step 1: Define Kw

K_w = [H⁺] [OH^-]

Step 2: ASSUME [H⁺] = [OH^-]

K_w = [H⁺]²

Step 3: Rearrange to find H+

√K_w = [H⁺]

Step 4: Put into pH

√2.09x10^-14 = [H⁺]

1.45x10^-7 = [H⁺]

pH = -log₁₀[1.45x10^-7 ]

pH = 6.84