a 5.29 g sample of impure sodiun chloride was dissolved and dihuted to a total volume of 250 0 ml. If 25.0 mL of the sodium chloride solation required 28.5 ml, of 0 300 M AgNO3, solution to reach the equivalence point, using chromate indicator, what was the percentage purity of the original sodiam chloride?
To find the percentage purity of the original sodium chloride, we need to determine the amount of sodium chloride in the impure sample and compare it to the amount of sodium chloride actually reacted with the AgNO3 solution.
Here's how we can calculate it:
1. Calculate the number of moles of AgNO3 used in the titration:
- Volume of AgNO3 solution used = 28.5 mL
- Concentration of AgNO3 solution = 0.300 M
- Moles of AgNO3 = (Volume in liters) x (Concentration in moles per liter)
- Convert the volume from mL to liters: 28.5 mL = 0.0285 L
- Moles of AgNO3 = 0.0285 L x 0.300 mol/L
2. Calculate the number of AgNO3 reacted with NaCl:
- According to the balanced chemical equation, 1 mole of AgNO3 reacts with 1 mole of NaCl.
- This means the moles of AgNO3 also represents the moles of NaCl that reacted.
3. Calculate the moles of NaCl in the original sample:
- Moles of NaCl in 25.0 mL = moles of AgNO3 used
- Moles of NaCl in total volume (250.0 mL) = (moles of NaCl in 25.0 mL) x (total volume/25.0 mL)
- Convert the total volume from mL to liters: 250.0 mL = 0.250 L
- Moles of NaCl in total volume = (moles of AgNO3 used) x (0.250 L / 0.025 L)
4. Calculate the percentage purity of the original NaCl:
- Percentage purity = (moles of NaCl in original sample / mass of original sample) x 100
- Mass of original sample = 5.29 g
Note: In this calculation, we are assuming that only NaCl is in the sample and no other impurities.
So, by plugging in the values into the equation, you should be able to find the percentage purity of the original sodium chloride.