Question

By titration it is found that 114.3 mL of 0.562 M NaOH is needed to titrate 48.0 mL of HCl. What is the concentration of the HCl in molarity? 

Answer 1

: NaOH +HCl ⇒NaCl + H2O.

So the ratio between the reactants is 1:1.

The moles of NaOH needed to neutralize the acid = volume of NaOH solution × concentration of NaOH solution

Moles of NaoH$= 0.1143 × 0.562 =0.0642 moles.$

Therefore 0.0642 mole of HCl is presented in the acid solution.

The concentration of HCl solution = moles of HCl/volume of HCl solution

Concentration of HCL $=\frac{0.0642}{0.048}=1.3375M$