A-Level pH, dissociation constants and buffers: worked solution

Question

A 250 cm³ buffer was made by mixing 120 cm³ of 0.100 mol dm⁻³ aqueous sodium hydroxide into 130 cm³ of valeric acid solution. The quantity of valeric acid still present in the 250 cm³ buffer was then found by titration. A 25.0 cm³ sample of the buffer needed 22.0 cm³ of 0.0500 mol dm⁻³ aqueous potassium hydroxide to react completely. Determine the amount, in moles, of valeric acid contained in the 250 cm³ buffer solution.

KOH neutralises the free valeric acid: acid + KOH -> salt + water (1:1). Moles KOH used = 0.0500 mol dm-3 x 22.0/1000 dm3 = 1.10 x 10⁻³ mol. So moles of valeric acid in the 25.0 cm3 sample = 1.10 x 10⁻³ mol. Scale up to the full 250 cm3 buffer: factor 250/25.0 = 10. Moles valeric acid in 250 cm3 = 1.10 x 10⁻³ x 10 = 1.10 x 10⁻² mol.