1- Find the number of moles of sodium hydroxide used.
# moles of NaOH = CC (mol. dm-3) x VOL (dm3)
# moles NaOH = 0.110 mol. dm-3 x 25.0 / 1000 dm3
# moles NaOH = 0.00275 mol NaOH (3 sig fig)
2- Using the mole ratio, calculate the # of moles of HCl present in the volume used.
The mole ratio is 1:1 since the formula for the neutralization reaction isHCL + NaOH ==>; H2O + NaCl
==> number of moles of HCl are also 0.00275
3- Find the concentration of the acid in the solution.
CC HCl = # moles HCl / Vol (dm3)
CC HCl = 0.00275 mol /(20.35/1000) dm3 ) = 0.135135135 M
We need to be consistent with the sig fig. in this case we should show 3 sigfig
the concentration would be then 0.135M