25 cm^3 of a solution of known 0.2M HCl is neutralised by titration by 21.5cm^3 of NaOH solution. Calculate then concentration of the NaOH solution to 3dp.

Recall n = c *(v/1000) if the volume is in Cm³.

Now, n(HCl)= 0.2 x (25.0/1000) 

n(HCl) = 5x10^-3

Since the relationship stoichiometrically is 1:1 (H⁺:OH^-), we can infer that 

n(NaOH) = 5x10^-3 

and by rearranging to c=n/(v/1000)

c(NaOH) = (5x10^-3) / (21.5/1000)

c(NaOH) = 10/43 moldm^-3 and to 3dp, c(NaOH) = 0.233moldm^-3