Question

To clean the impeller blades of a stainless steel tank, plant technicians pour 210.0 L of a 240.0 g HNO3/L solution in a 820.0-L tank to soak. When the tank needed to be emptied, the tank was first filled with water entering at a rate of 1.90 L/s. (Figure 1). At t = tF, the tank is completely full of solution. At this point water continued to enter the tank at the same rate, but solution was also drained from the bottom of the tank at 1.90 L/s (Figure 2). Make the assumption that the fluid density does not change and answer the following questions. 1.) What is the value of tF, the time it takes to fill the tank? 2.) What is CF, the concentration of HNO3 in the tank at tF? 3.)At what point after starting to add water to the tank is the concentration of HNO3 0.0500 of the original concentration?

Answer 1

The capacity of the tank is 820 L

It has HNO₃ solution 210 L. So capacity remaining is 610 L.

The tank was first filled with water entering at a rate of 1.90 L/s.

At t = tF, the tank is completely full so to fill 610 L at 1.9 L/ s we will need

610L/1.9 L/s = 321 seconds or 5.35 minutes

The concentration of HNO3 when the tank is full is

240 g/ L x 210 L = 50400 g of HNO3 total now diluted to 820 L

so 50400 g in 820 L is 61.46 g/L HNO3

240 g/L is the original concentration

0.0500 of the original is 240 g/L x 0.050 = 12 g/L

The concentration will be 12g/L since there is no change in volume of the tank the total HNO3 will be 12g/L x 820L = 9840 g of HNO3

50400g - 9840 g = 40560 g of HNO3 will have been removed.

1.9 L/s x 61.46 g/L every second 116.77 g of HNO3 is being removed so to remove 40560 g of HNO3 you will need

40560/116.77 = 347 seconds

So after 347 seconds + 321 seconds = 668 seconds later the concentration will be 0.050 of the original.