Naqueous 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

Fine out your self ponk

Reaction type : double replacement hcl + lioh = h2o + licl