25 cm3 of NaOH was titrated with 0.050 mol dm-3 HCl. NaOH + HCl --> NaCl + H2O. 21.5 cm3 HCl neutralised 25 cm3 NaOH. Concentration of NaOH in mol dm-3?
21.5 cm3 / 1000 = 0.0215 dm3 of HCl. n = cv so (0.050) x (0.0215) = 0.001075 mols of HCl. NaOH + HCl --> NaCl + H2O so there is a 1:1 ratio of reactants to products. Therefore there are 0.001075 mols of NaOH. n = cv so n /v = c. (0.001075) / (0.025) = 0.043 mol dm-3 is the concentration of NaOH
AS
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