Formula to find OH??

'As the pH of the solution is increased from 8 to 10 two phenolic protons are removed, a lactone ring opens and a pink compound is formed, H2P P2. This reaction is fast at room temperature and is widely used in acid-base titrimetry to indicate end-points, as you well know (from 1st year chemistry). Unfortunately, at higher pHs the pink colour slowly fades representing a further structural change: P2 + OH POH3 The object of this experiment is to determine the kinetics of this slow reaction, whose rate has been shown to be equal to k2[P2][OH].'

250.0 ml of a ~0.300 mol dm-3 solution of NaCl

75.0 ml of stock NaOH + the balance NaCl;

50.0 ml of NaOH + NaCl

25.0 ml NaOH + the balance NaCl.

Concentration of NaOH not guve just labelled 'Stock'

How do I get OH values for my equation k2 = k1/[OH ]0 ??

My k1 =

0.005

0.0093

0.0261

0.0049