A solution was made by dissolving 1.5g of sodium hydroxide pellets containing inert impurities in water and making it to 250cm³ of solution.If 20cm³ of this solution is neutralized exactly by 13cm³ of 1M of hydrochloric acid, calculate the percentage purity of sodium hydroxide
To solve this problem, we need to first determine the number of moles of hydrochloric acid used in the neutralization reaction.
The volume of hydrochloric acid used is 13 cm³, and the concentration of the hydrochloric acid is 1M. This means that there is 1 mole of hydrochloric acid in 1 liter of solution (1000 cm³). Therefore, there are 13/1000 = 0.013 moles of hydrochloric acid in 13 cm³.
According to the balanced chemical equation for the neutralization reaction between sodium hydroxide and hydrochloric acid, 1 mole of sodium hydroxide reacts with 1 mole of hydrochloric acid.
Since 20 cm³ of the sodium hydroxide solution is exactly neutralized by 13 cm³ of hydrochloric acid, we can conclude that 0.013 moles of sodium hydroxide are present in 20 cm³ of the solution.
To find the moles of sodium hydroxide dissolved in the entire 250 cm³ solution, we can set up a proportion:
(0.013 moles / 20 cm³) = (x moles / 250 cm³)
Simplifying the proportion, we find:
x moles = (0.013 moles / 20 cm³) * 250 cm³
x moles = 0.01625 moles
Now we can calculate the mass of sodium hydroxide dissolved in the solution:
mass of sodium hydroxide = moles * molar mass
mass of sodium hydroxide = 0.01625 moles * 40 g/mole (the molar mass of sodium hydroxide)
mass of sodium hydroxide = 0.65 g
Finally, we can calculate the percentage purity of sodium hydroxide:
percentage purity = (mass of pure sodium hydroxide / mass of impure sodium hydroxide) * 100
percentage purity = (0.65 g / 1.5 g) * 100
percentage purity = 43.33%
Therefore, the percentage purity of the sodium hydroxide pellets is approximately 43.33%.