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%.
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
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