If ten drops of 0.3 M HCl are mixed with ten drops of 0.3 M Na2S2O3, what will be the concentration of each chemical in the mixture?
Please help, i am pretty sure that the concentration should be 0.3 M but i am not sure if i am supposed to do somthing with the ten drops.
Thank you for any help
For clarity, just assign some number to the volume of a drop. Actually, a drop is between 0.03 and 0.05 mL (depending upon the size of the opening AND on the material that the buret/pipet/dropper is made). BUT, let's just make things simple by saying a drop is 1 mL. We know that isn't true but it doesn't matter as long as we are consistent.
So 0.3 M x 10 mL(0.01 L) = 0.003 mols HCl.
and 0.3 M x 0.01 L = 0.003 mols Na2S2O3.
Molarity HCl = #mols HCl/L solution = 0.003/0.02 L = 0.15 M. (The volume os 10 mL + 10 mL = 20 mL = 0.02 L)
Molarity Na2S2O3 = 0.15 by the same reasoning.
Of course, we should have know, just basically, that 0.15 M is the answer.
We add 10 drops 0.3 to 10 drop 0.3 so each 0.3 has been diluted by a factor of 2 so the final concentration is 0.3/2 = 0.15. I hope this is clear.
thank you very much
To determine the concentration of each chemical in the mixture, we need to consider the number of moles of each chemical present in the solution.
First, we need to determine the number of moles of HCl and Na2S2O3 in the ten drops of each solution.
Given:
Number of drops = 10
Concentration of HCl = 0.3 M
Concentration of Na2S2O3 = 0.3 M
To calculate the number of moles, we can use the formula:
Moles (mol) = Concentration (M) x Volume (L)
The volume of one drop is usually considered negligible, so we can assume that the volume of 10 drops is also negligible. Therefore, the concentration remains the same.
Number of moles of HCl in 10 drops:
Moles of HCl = 0.3 M x 10 drops = 3.0 x 10^-3 mol
Similarly, the number of moles of Na2S2O3 in 10 drops is also:
Moles of Na2S2O3 = 0.3 M x 10 drops = 3.0 x 10^-3 mol
Now that we know the number of moles for each chemical, we can calculate the concentration in the mixture.
To determine the total volume of the mixture, we need to consider the volumes of both solutions.
Given that the volume of one drop is negligible, the total volume of the mixture will be approximately 10 drops.
Concentration of HCl in the mixture:
Concentration of HCl = Moles of HCl / Total Volume of Mixture
= 3.0 x 10^-3 mol / 10 drops ≈ 0.3 M
Concentration of Na2S2O3 in the mixture:
Concentration of Na2S2O3 = Moles of Na2S2O3 / Total Volume of Mixture
= 3.0 x 10^-3 mol / 10 drops ≈ 0.3 M
Therefore, in the mixture, both HCl and Na2S2O3 will have a concentration of approximately 0.3 M.
To determine the concentration of each chemical in the mixture, we need to calculate the total volume of the mixture first. Assuming that each drop is approximately 0.05 mL (a common approximation), the total volume of the mixture is:
Total volume = Number of drops * Volume per drop
Total volume = 10 drops * 0.05 mL/drop
Total volume = 0.5 mL
Now, we can use this total volume to find the concentration of each chemical in the mixture.
For HCl:
The concentration of HCl is given as 0.3 M, which means there are 0.3 moles of HCl in 1 liter of solution. To find the number of moles in the mixture, we can use the equation:
moles of HCl = concentration of HCl * volume of solution
moles of HCl = 0.3 M * 0.5 L
Since the total volume of the mixture is 0.5 mL (or 0.0005 L), we can substitute this value into the equation:
moles of HCl = 0.3 M * 0.0005 L
moles of HCl = 0.00015 moles
Since we have the number of moles, we can calculate the concentration of HCl in the mixture as follows:
concentration of HCl in the mixture = moles of HCl / volume of the mixture
concentration of HCl in the mixture = 0.00015 moles / 0.0005 L
concentration of HCl in the mixture = 0.3 M
Therefore, the concentration of HCl in the mixture is 0.3 M.
For Na2S2O3:
Using the same approach, we can calculate the concentration of Na2S2O3 in the mixture. The concentration of Na2S2O3 is also given as 0.3 M, so the number of moles can be calculated as follows:
moles of Na2S2O3 = concentration of Na2S2O3 * volume of solution
moles of Na2S2O3 = 0.3 M * 0.5 L
moles of Na2S2O3 = 0.00015 moles
To calculate the concentration of Na2S2O3 in the mixture:
concentration of Na2S2O3 in the mixture = moles of Na2S2O3 / volume of the mixture
concentration of Na2S2O3 in the mixture = 0.00015 moles / 0.0005 L
concentration of Na2S2O3 in the mixture = 0.3 M
Therefore, the concentration of Na2S2O3 in the mixture is 0.3 M, similar to the concentration of HCl.