If I add water to 100 mL of a 0.15 M NaOH solution until the final volume is 150 mL, what will the molarity of the diluted solution be?

Question

Jeremiah Adams · Accepted Answer

Approx. #0.10# #mol*L^-1# #"Concentration"# #=# #("MOLES of SOLUTE")/("Volume of SOLUTION")# So all we need to is to calculate the one quantity; #"Volume of SOLUTION"# has been specified to be #150# #mL#. So, #"MOLES of SOLUTE"# #=# #0.15*mol*L^-1# #xx# #100xx10^-3L# #=# #??"mol"#; this was our starting solution. And final #"CONCENTRATION"# #=# #(0.15*mol*cancel(L^-1)xx100xx10^-3cancel(L))/(150xx10^-3*L)# #=# #??mol*L^-1#

Wyatt Atkins · Answer

undefined The molarity of the diluted solution can be calculated using the formula:

$$ M_1 \times V_1 = M_2 \times V_2 $$

Where:
$ M_1 = 0.15 \, \text{M} $ (initial molarity)
$ V_1 = 100 \, \text{mL} $ (initial volume)
$ V_2 = 150 \, \text{mL} $ (final volume)
$ M_2 $ is the unknown molarity of the diluted solution.

Rearranging the formula to solve for $ M_2 $:

$$ M_2 = \frac{M_1 \times V_1}{V_2} $$

$$ M_2 = \frac{0.15 \times 100}{150} $$

$$ M_2 = \frac{15}{150} $$

$$ M_2 = 0.1 \, \text{M} $$

So, the molarity of the diluted solution will be 0.1 M.