Question

An impure sample of potassium hydrogen phthalate was titrated to determine the percent composition KHP. What is the percentage of potassium hydrogen phthalate if 2.81 g of the mixture required 35.61 mL of 0.152 mol/L NaOH to reach the end point

Answer 1

For an impure sample of potassium hydrogen phthalate, the percentage of potassium hydrogen phthalate is

m'=39.3%

What is the percentage of potassium hydrogen phthalate if 2.81 g of the mixture required 35.61 mL of 0.152 mol/L NaOH to reach the end point?

Generally, the equation for the mol base  is mathematically given as

mol base = M*V

mol base = 0.152*35.61

mol base = 5.41272 mmol

In conclusion

mass = mmol/1000*MW

m== 5.41272 /1000*204.22

m= 1.1053 g of KHP

Hence, m'=39.3%

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