For preparing 3.00 L of 1 M NaOH by mixing portions of two stock solutions ( A and B ) of 2.50 M NaOH and 0.40 M NaOH respectively. Find out the amount of B stock solution (in L ) added.

Step-by-step solution

Moles needed =(3.00 ~L )(1.00 M )=3 ~mol . Suppose that x ~L of 2.50 M NaOH added, then (3-x) L of 0.40 M NaOH added. The number of moles of solute from the more concentrated solution is 2.50 , that become less concentrated solution is (0.40)(3.00-x) . The total number of moles is 3.00 . (2.50 x)+(0.40)(3.00-x)=3.00 or 2.10 x+1.2=3 Or 2.10 x=1.8 or x=0.857 ~L of 2.50 M NaOH (3-0.857)=2.14 L of 0.40 M NaOH