Question

at 25 C, a solution is prepared by adding 50 ml of 0.2 M NaOH to 75 ml of 0.1 M NaOH. What is the total concentratoin of hydorxide ion in the solution? Assume volumes are additive.

Answer 1

Formula to be used:

n=M*V

Where: n is number of moles of NaOH

            M=molarity

            V is volume

From 50 ml of 0.2 M NaOH:

M=0.2 M

V= 50 ml =0.05 L

n=M*V

=0.2*0.05

=0.01 mol

From 75 ml of 0.1 M NaOH:

M=0.1 M

V= 75 ml =0.075 L

n=M*V

=0.1*0.075

=0.0075 mol

So total number of moles of NaOH = 0.01+0.0075 = 0.0175 mol

NaOH ---->Na+ OH-

so, number of moles of hydroxide ion (OH-) = number of moles of NaOH = 0.0175 mol

Total volume of mixture = 50 ml +75 ml =125 ml =0.125 L

Now use:

M=n/V

here:

M= molarity or concentration of Hdroxide ion

n= number of moles of Hdroxide ion = 0.0175 mol

V=volume of solution = 0.125 L

so, M=n/V

      = 0.0175/ 0.125

      = 0.14 M

Answer: 0.14 M

Hope I have explained nicely for you to understand