One of the more common agents for standardizing solutions of potassium hydroxide
ID: 794294 • Letter: O
Question
One of the more common agents for standardizing solutions of potassium hydroxide (KOH) is potassium acid phtalate (KHC8H4O4), or, more accurately, potassium hydrogen phthalate. From the following data, determine the average molarity of a KOH solution that was standardized with KHP.
Trial A: 45.12 mL of KOH solution neutralized 0.2573 g of KHP
Trial B: 48.89 mL of KOH solution neutralized 0.2250 g of KHP
Trial C: 45.50 mL of KOH solution neutralized 0.2502 g of KHP
If someone could show the work so I can understand this that would be great, thanks!
Explanation / Answer
KOH + KHC8H4O4 => K2C8H4O4 + H2O
(1) Moles of KOH = Moles of KHC8H4O4 = mass/molar mass of KHC8H4O4
= 0.2573/204.22 = 0.0012599 mol
Molarity = moles/volume of KHP
= 0.0012599/0.04512 = 0.02792 M
(2) Moles of KOH = Moles of KHC8H4O4 = mass/molar mass of KHC8H4O4
= 0.2250/204.22 = 0.0011018 mol
Molarity = moles/volume of KHP
= 0.0011018/0.04889 = 0.02254 M
(3) Moles of KOH = Moles of KHC8H4O4 = mass/molar mass of KHC8H4O4
= 0.2502/204.22 = 0.0012251 mol
Molarity = moles/volume of KHP
= 0.0012251/0.04550 = 0.02693 M
Average molarity = (0.02792 + 0.02254 + 0.02693)/3
= 0.02580 M
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