Question

A holding tank with capacity 5000 litres initially contains 1500 litres of 25 mM NaCl solution. a) What is the final concentration of NaCl (expressed in mM) if an additional 3000 litres of 25 mM NaCl solution is pumped into the tank? b) Instead of (a), if 3000 litres of water are added to the tank, what is the final concentration of NaCl (expressed in mM)? c) Instead of (a) or (b), if an additional 500 litres of 25 mM NaCl solution plus 3000 litres of water are added, what is the final concentration of NaCl:

iv. Expressed as ppm? (ppm means g per 106 g of solution) You can assume that the density of the solution (NaCl + water) is the same as the density of water – this assumption is appropriate since this is a dilute solution. Assume density of water to be 1 g/cm3 .

Note: mM means milli-molar (1 mM means 1 milli-mol/liter)

Answer 1

a)
final concentration= (M1*V1+ M2*V2) /(V1+V2)
                                      = (25*1500 + 25*3000) / (1500+3000)
                                        = 25 mM

b)
Concentration of NaCl in pure water =0
so, M2 will be 0 here
final concentration= (M1*V1+ M2*V2) /(V1+V2)
                                      = (25*1500 + 0*3000) / (1500+3000)
                                        = 8.33 mM

c)
final concentration= (M1*V1+ M2*V2+M3V2) /(V1+V2+V3)
                                      = (25*1500 + 25*500 + 0*3000) / (1500+500+3000)
                                        = 10 mM

d)
we need convert 10mM in ppm
10 mM means 10mmol of NaCl in 1 L solution.
Mass 0f solution = density * volume
                       = (1000 g/L) * 1L
                        = 1000 g

Molar mass of NaCl = 58.5 gm
mass of Nacl = number of moles *molar mass
                           = 0.01 * 58.5
                            = 1.71*10^-4

ppm= mass of solute * 10^6 / total mass
           = (1.71*10^-4)*10^6   /   (1000)
             = 0.171
Answer: 0.171 ppm