A #1.5*dm^3# volume of #0.30*mol*dm^-3# #NaCl(aq)# was mixed with a #2.5*dm^3# volume of #0.70*mol*dm^-3# #NaCl(aq)#. What is the final concentration of #NaCl(aq)#?

Question

A #1.5*dm^3# volume of #0.30*mol*dm^-3# #NaCl(aq)# was mixed with a #2.5*dm^3# volume of #0.70*mol*dm^-3# #NaCl(aq)#.  What is the final concentration of #NaCl(aq)#?

Cameron Andrews · Accepted Answer

#[NaCl(aq)]=0.55*mol*L^-1#.

#"Concentration"# #=# #"Moles of solute"/"Volume of solution"#. We need to figure out how many ions are in the solution overall and divide that number by the volume. #"Moles of NaCl(i)"# #=# #1.50*dm^3xx0.30*mol*dm^-3=0.45*mol# #"Moles of NaCl(ii)"# #=# #2.50*dm^3xx0.70*mol*dm^-3=1.75*mol# And thus #[NaCl]# #=# #(0.45*mol+1.75*mol)/(4.0*dm^3)# #=# #0.55*mol*dm^-3#. Note that #1*dm^3# #=# #(10^-1m)^3# #=# #10^-3m^3=1L#. We have (reasonably) assumed the volumes of solution to be additive.

Kai Bowman · Answer

undefined To find the final concentration of NaCl(aq) after mixing the two solutions, you can use the formula:

$$C_f = \dfrac{C_1V_1 + C_2V_2}{V_1 + V_2}$$

Where:
$C_f$ = final concentration
$C_1$ = concentration of the first solution
$C_2$ = concentration of the second solution
$V_1$ = volume of the first solution
$V_2$ = volume of the second solution

Given:
$C_1 = 0.30 \, \text{mol dm}^{-3}$
$C_2 = 0.70 \, \text{mol dm}^{-3}$
$V_1 = 1.5 \, \text{dm}^3$
$V_2 = 2.5 \, \text{dm}^3$

Plugging in the values:

$$C_f = \dfrac{(0.30 \, \text{mol dm}^{-3})(1.5 \, \text{dm}^3) + (0.70 \, \text{mol dm}^{-3})(2.5 \, \text{dm}^3)}{1.5 \, \text{dm}^3 + 2.5 \, \text{dm}^3}$$

$$C_f = \dfrac{(0.45 \, \text{mol}) + (1.75 \, \text{mol})}{4 \, \text{dm}^3}$$

$$C_f = \dfrac{2.20 \, \text{mol}}{4 \, \text{dm}^3}$$

$$C_f = 0.55 \, \text{mol dm}^{-3}$$

Therefore, the final concentration of NaCl(aq) is $0.55 \, \text{mol dm}^{-3}$.