Question

Calculate the concentration (in molarity) of NaOH solution, if 25.0mL of the solution are needed to neutralize 17.4 mL of a 0.132M HCL solution.

Answer 1

There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. For this we need the neutralization reaction of HCl and NaOH which is:

HCl + NaOH = NaCl + H2O

Calculation is as follows:

Molarity of NaOH = (0.132 M HCl)(.0174 L)(1 mol NaOH / 1 mol HCl ) / .025 L solution = 0.1 mol NaOH / L = 0.1 M

Answer 2

Explanation:

The given data is as follows.

      $V_{1}$ = 25 ml,     $V_{2}$ = 17.4 ml

      $M_{1}$ = ?,          $M_{2}$ = 0.132 M

Therefore, calculate molarity of the solution as follows.

         $M_{1}V_{1} = M_{2}V_{2}$

        $M_{1} \times 25 ml = 0.132 M \times 17.4 ml$

           $M_{1}$ = 0.091 M

Thus, we can conclude that the concentration (in molarity) of NaOH solution is 0.091 M.