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A 0.683 g impure sample of strontium hydroxide is dissolved in 120 mL of 0.1010 M HCl. Titration of the excess with 0.1020 M NaOH required 36.6 mL to reach the equivalence point. Calculate the percent purity of the sample.

Answer :

The percent purity of strontium hydroxide in the mass of the given sample is 74.7%.

What is the percent purity of a substance?

The percent purity of a substance is the ratio of the actual amount of a substance present in a sample and the ratio of the whole amount of the sample.

  • Percent purity = (mass of pure sample/ mass of impure sample) * 100%

Equation of reaction: Sr(OH)â‚‚ + 2 HCl ---> SrClâ‚‚ + 2 Hâ‚‚O

The mass of the pure strontium hydroxide is calculated as follows:

moles of initial HCl sample = 0.1010 * 120 mL * 1 L / 1000 mL

moles of initial HCl sample = 0.01212 moles

Moles of HCl that reacted with NaOH = 0.1020 * 36.6 * 1 L / 1000 mL

Moles of HCl that reacted with NaOH = 0.00373 moles

moles of HCl that reacted with strontium hydroxide = 0.01212 - 0.00373

moles of HCl that reacted with strontium hydroxide = 0.00839 moles

Moles of strontium hydroxide that reacted with HCl = 0.00839/2 moles

Mass of strontium hydroxide that reacted with HCl = 0.00839/2 moles * 121.63 g/mol

Mass of strontium hydroxide that reacted with HCl = 0.510 g

Percent purity = 0.510/0.683 * 100 %

Percent purity = 74.7%

Learn more about percent purity at: https://brainly.com/question/17021926

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Rewritten by : Jeany

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