25cm3 NaOH was titrated with 0.05mol dm-3 HCl. 21.5m3 of HCl neutralised 25cm3 of NaOH. What is the concentration of NaOH in mol dm-3?
NaOH + HCl --> NaCl + H2O 1 : 1 : 1 moles (mol) = concentration (mol dm-3) x volume (dm3) n(HCl) = 0.05 (mol dm-3) x 21.5/1000 (dm3) = 0.001075 moles (1.075 x 10-3) Using 1:1 molar ratio: n(NaOH) = n(HCl) n(NaOH) = 0.001075 moles concentration (mol dm-3) = moles (mol) / volume (dm3) [NaOH] = 0.001075 (mol) / 0.025 (dm3) = 0.043 mol dm-3
BA
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