If a chemist titrates 150.0ml of NaOH with a 10.0 M solution of HCl and requires 83.4ml of the acid to reach the endpoint, what is the concentration of the NaOH?
Since we are including the equivalence point due to the 83.4 mL of the acid being added, you know that moles of acid = moles of base at Equivalence point.
Convert volume of acid to Liters. 83.4 mL = 0.0834 L. Multiply Molarity of acid by Liters to get moles.10.0 mol/L x 0.0834=0.834 mol HCL. 0.834 mol of NaOH exist at eq point, therefore 150.0 mL = 0.1500 L, and 0.834 mol NaOH/0.1500L = 5.56 M NaOH