A sodium hydroxide sample is contaminated with sodium chloride. A 0.240 gram sample of this impure sample is dissolved in water and titrated with 21.50 mL of 0.130 M H2C2O4. Calculate the percent sadium hydroxide in the impure sample [(mass of NaOH in impure sample/mass of impure sample)*100%].
Please help me
2NaOH + H2C2O4 ==> Na2C2O4 + 2H2O
mols H2C2O4 = M x L = ?
Using the coefficients in the balanced equation, convert mols H2C2O4 to mols NaOH (that's mols NaOH = 2 x mols H2C2O4).
Then g NaOH = mols NaOH x molar mass NaOH.
Finally,
(g NaOH/mass sample)*100 = %NaOH
To calculate the percent sodium hydroxide in the impure sample, we need to determine the amount of sodium hydroxide present in the solution.
First, let's determine the number of moles of H2C2O4 used in the titration:
moles of H2C2O4 = concentration of H2C2O4 x volume of H2C2O4
= 0.130 M x 0.0215 L
= 0.002795 moles
According to the balanced equation between H2C2O4 and NaOH, the mole ratio is 1:2. Therefore, there must be twice as many moles of NaOH present.
moles of NaOH = 2 x moles of H2C2O4
= 2 x 0.002795 moles
= 0.00559 moles
Now, we need to find the mass of the impure sample used in the titration. Since the impure sample is a mixture of sodium chloride and sodium hydroxide, the mass of sodium hydroxide can be calculated by subtracting the mass of sodium chloride from the mass of the impure sample.
mass of impure sample = 0.240 grams
Next, we need to calculate the mass of sodium chloride present in the impure sample. To do this, we need to consider the molar mass of sodium chloride (58.44 g/mol) and the mole ratio between sodium chloride and sodium hydroxide (1:1).
moles of NaCl = moles of NaOH = 0.00559 moles
mass of NaCl = moles of NaCl x molar mass of NaCl
= 0.00559 moles x 58.44 g/mol
= 0.3267 grams
Now, we can calculate the mass of sodium hydroxide in the impure sample by subtracting the mass of sodium chloride from the mass of the impure sample:
mass of NaOH in impure sample = mass of impure sample - mass of NaCl
= 0.240 grams - 0.3267 grams
= -0.0867 grams (negative because we have more chloride than the sample)
Since we cannot have a negative mass, it means there was no sodium hydroxide in the impure sample. Therefore, we cannot determine the percent sodium hydroxide in the impure sample based on the given information.
To solve this problem, you need to set up and solve an equation based on the stoichiometry of the reaction between sodium hydroxide (NaOH) and oxalic acid (H2C2O4). Here's how you can approach it:
1. Convert the volume of the oxalic acid solution used in the titration to moles. You can use the formula:
Moles of H2C2O4 = Concentration (M) x Volume (L)
In this case, the concentration is 0.130 M, and the volume is 21.50 mL (convert it to liters by dividing by 1000).
2. Determine the stoichiometry of the reaction. The balanced chemical equation between NaOH and H2C2O4 is:
2 NaOH + H2C2O4 -> Na2C2O4 + 2 H2O
This means that for every 2 moles of NaOH, you need 1 mole of H2C2O4.
3. Use the stoichiometry to determine the moles of NaOH consumed in the reaction. Divide the moles of H2C2O4 obtained in step 1 by the stoichiometric ratio of NaOH to H2C2O4 (which is 2:1 in this case). This will give you the moles of NaOH.
4. Calculate the mass of NaOH in the impure sample. Use the moles of NaOH from step 3 and the molar mass of NaOH (which is 22.99 g/mol) to calculate the mass.
Mass of NaOH = Moles of NaOH x Molar mass of NaOH
5. Calculate the percent NaOH in the impure sample. Use the mass of NaOH from step 4 and the mass of the impure sample (which is 0.240 g) to calculate the percentage.
Percent NaOH = (Mass of NaOH / Mass of impure sample) x 100%
Follow these steps to solve the problem and find the percent of sodium hydroxide in the impure sample.