I used 10mL of unknown NaCl-NaBr mixture titrate with 0.10M silver nitrate solution.
10.15mL of silver nitrate solution is used.
the conc. of NaCl-NaBr mixture is 8g/dm3
I need to calculate the % composition of the NaCl-NaBr mixture.
I used the following calculation steps but i can't get the ans.
a + b = c......(1), where c is the weight of the mixture used, consisting a gram of NaCl and b gram of NaBr.
c= 8g x 10/1000 = 0.08g
no of moles of Cl react= a/58.44
no of moles of Br react= b/102.9
no of moles of Ag react=(0.1 x 10.15)/1000= total no of moles of Cl and Br react=a/58.44 + b/102.9
therefore, (0.1 x 10.15)/1000=a/58.44 + b/102.9.....(2)
values of a and b should easily be found by solving equation (1) and (2), however I am fail to do so.
What's wrong with my calculation?!?!
See my response to your first post below. It looks the same as your procedure. If you want to post your work I will try to find the error.
Your calculation seems to be on the right track, but there are a couple of errors in the equations you are using.
Let's go through the calculations step by step to find the correct solution.
1. First, calculate the mass of the mixture used:
Mass of mixture = Concentration of mixture x Volume of mixture
Mass of mixture = 8 g/dm³ x 10 mL / 1000 (conversion from mL to dm³)
Mass of mixture = 0.08 g
2. Set up the equation to relate the weights of NaCl and NaBr in the mixture:
a + b = 0.08 g
3. Next, determine the moles of Cl that react with silver nitrate:
Moles of Cl = mass of NaCl / molar mass of NaCl
Moles of Cl = a / 58.44 g/mol
4. Similarly, calculate the moles of Br that react with silver nitrate:
Moles of Br = mass of NaBr / molar mass of NaBr
Moles of Br = b / 102.9 g/mol
5. The moles of Ag that react with a mixture of NaCl and NaBr can be calculated using the volume and concentration of silver nitrate:
Moles of Ag = concentration of AgNO₃ x volume of AgNO₃ / 1000 (conversion from mL to dm³)
Moles of Ag = 0.10 mol/dm³ x 10.15 mL / 1000
Moles of Ag = 0.01015 mol
Now, the correct equation relating these quantities can be set up:
0.01015 mol = (a / 58.44 g/mol) + (b / 102.9 g/mol)
You can rearrange this equation to solve for one variable in terms of the other. Let's solve it for b:
b = (0.01015 mol - (a / 58.44 g/mol)) x (102.9 g/mol)
Now you have a relation between a and b. You can substitute this equation back into equation (1) to solve for the composition of the mixture.
a + [(0.01015 mol - (a / 58.44 g/mol)) x (102.9 g/mol)] = 0.08 g
Simplify and solve this equation to find the value of a. Once you have a, you can substitute it back into equation (1) to calculate the value of b.
Finally, to calculate the percentage composition of NaCl and NaBr in the mixture:
% NaCl = (mass of NaCl / mass of mixture) x 100
% NaBr = (mass of NaBr / mass of mixture) x 100
By substituting the calculated values of a and b into these equations, you can determine the percentage composition of the NaCl-NaBr mixture.