- 0.05 L needs to be added to the original 0.12 L solution in order to dilute it from 0.13 M to 0.23 M.
The dilution problem uses the equation :
[tex]M_aV_a= M_bV_b[/tex]
The initial molarity (concentration) [tex]M_a =[/tex] 0.13 M
The initial volume [tex]V_a[/tex] = 0.12 L
The desired molarity (concentration) [tex]M_b[/tex] = 0.23 M
The volume of the desired solution [tex]V_b[/tex] = ( 0.12 + x L )
Substituting values in above equation;
(0.13 M ) (0.12 L) = (0.23 M ) (0.12 L + x L)
0.0156 M L = 0.0276 M L + 0.23 x M L
- 0.012 M L = 0.23 x M L
x = - 0.05
Therefore, - 0.05 L needs to be added to the original 0.12 L solution in order to dilute it from 0.13 M to 0.23 M.
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