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
WF
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