Question

please help with 3 & 4! thank you

3. A solution of KOH was standardized using KHP. A 6.245 g sample of KHP required 41.05 mL KOH in one titration and 40.85 mL in the second. Calculate the average molarity of the KOH solution.

Answer 1

3) The KOH concentration is calculated:

[KOH] 1 = g KHP / MM KHP * V KOH = 6.245 g / 204.22 g / mol * 0.04105 L = 0.745 M

Trial 2: 0.749 M

Average = 0.747 M

4) The reaction that occurs is:

Ca (OH) 2 + 2 HKP = Ca (KP) 2 + 2 H2O

The molarity of Ca (OH2) is calculated:

[Ca (OH) 2] = 2.5 g / 204.22 g / mol * 0.020 L * 2 = 0.306 M

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