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A student prepared a standard solution of HCl to standardize. 0.1423 g of anhydrous soda ash was dissolved in 25 ml distilled H2O, and 2 gtts of methyl orange TS. Upon titration, there is 23.18 ml left in 25 ml of standard solution in the burette until pink color is observed. Determine the n of the solution
Accepted Answer
Normality (N) of the HCl solution:
Step 1: Calculate the moles of Na2CO3 in the sample
-Mass of Na2CO3 = 0.1423 g
-Molar mass of Na2CO3 = 105.99 g/mol
Moles of Na2CO3 = 0.1423 g / 105.99 g/mol = 0.001343 mol
Step 2: Determine the moles of HCl that reacted with Na2CO3
- According to the balanced chemical equation:
2HCl + Na2CO3 → 2NaCl + H2O + CO2
- 1 mole of Na2CO3 reacts with 2 moles of HCl
Moles of HCl = 2 x Moles of Na2CO3 = 2 x 0.001343 mol = 0.002686 mol
Step 3: Calculate the volume of HCl solution used for titration
- Volume of HCl solution in burette = 25.00 mL
- Volume of HCl solution left after titration = 23.18 mL
Volume of HCl solution used = 25.00 mL - 23.18 mL = 1.82 mL
Step 4: Calculate the Normality of the HCl solution
- Normality = (Moles of HCl) / (Volume of HCl solution in liters)
- Convert volume of HCl solution to liters:
Volume in liters = 1.82 mL / 1000 mL/L = 0.00182 L
- Normality = 0.002686 mol / 0.00182 L = 1.476 N
Therefore, the Normality (N) of the HCl solution is 1.476 N.
Step 1: Calculate the moles of Na2CO3 in the sample
-Mass of Na2CO3 = 0.1423 g
-Molar mass of Na2CO3 = 105.99 g/mol
Moles of Na2CO3 = 0.1423 g / 105.99 g/mol = 0.001343 mol
Step 2: Determine the moles of HCl that reacted with Na2CO3
- According to the balanced chemical equation:
2HCl + Na2CO3 → 2NaCl + H2O + CO2
- 1 mole of Na2CO3 reacts with 2 moles of HCl
Moles of HCl = 2 x Moles of Na2CO3 = 2 x 0.001343 mol = 0.002686 mol
Step 3: Calculate the volume of HCl solution used for titration
- Volume of HCl solution in burette = 25.00 mL
- Volume of HCl solution left after titration = 23.18 mL
Volume of HCl solution used = 25.00 mL - 23.18 mL = 1.82 mL
Step 4: Calculate the Normality of the HCl solution
- Normality = (Moles of HCl) / (Volume of HCl solution in liters)
- Convert volume of HCl solution to liters:
Volume in liters = 1.82 mL / 1000 mL/L = 0.00182 L
- Normality = 0.002686 mol / 0.00182 L = 1.476 N
Therefore, the Normality (N) of the HCl solution is 1.476 N.
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