N aqueous solution of sodium hydroxide is standardized by titration with a 0.154 m solution of hydrochloric acid. if 17.5 ml of base are required to neutralize 17.6 ml of the acid, what is the molarity of the sodium hydroxide solution?

The balanced equation for the above reaction is as follows;

NaOH + HCl - - > NaCl + H₂O

stoichiometry of NaOH to HCl is 1:1

Number of HCl moles reacted - 0.154 mol/L x 0.0176 L = 0.00271 mol

the number of NaOH moles reacted = number of HCl moles reacted

number of NaOH moles reacted - 0.00271 mol

number of NaOH moles in 17.5 mL - 0.00271 mol

therefore NaOH moles in 1000 mL - 0.00271 mol / 17.5 mL x 1000 mL

molarity of NaOH - 0.155 M