A student is asked to standardize a solution of calcium hydroxide. He weighs out

Question

A student is asked to standardize a solution of calcium hydroxide. He weighs out 0.924 g potassium hydrogen phthalate (KHC_8H_4O_4. treat this as a monoprotic acid). It requires 20.5 mL of calcium hydroxide to reach the endpoint. A. What is the molarity of the calcium hydroxide solution? M This calcium hydroxide solution is then used to titrate an unknown solution of perchloric acid. B. If 22.6 mL of the calcium hydroxide solution is required to neutralize 13.4 mL of perchloric acid, what is the molarity of the perchloric acid solution? M A student is asked to standardize a solution of barium hydroxide. He weighs out 0.924 g potassium hydrogen phthalate (KHC_8H_4O_4, treat this as a monoprotic acid). TL requires 25.4 mL of barium hydroxide to reach the endpoint. A. What is the molarity of the barium hydroxide solution? M This barium hydroxide solution is then used to titrate an unknown solution of hydroiodic acid. B. If 28.1 mL of the barium hydroxide solution is required to neutralize 26.0 mL of hydroiodic acid, what is the molarity of the hydroiodic acid solution? M

Explanation / Answer

1.KHP molecular weight = 204.23 g/mole
0.924 g / 204.23 g/mole = 0.00452 mole
Ca(OH)2 + 2KHP --> 2H2O + Ca(KP)2
0.00452 / 2 = 0.00226 moles of Ca(OH)2 will neutralize 0.00452 moles KHP
0.0205 L x M = 0.00226 mole
Ca(OH)2 = 0.110 M

22.6 mL x 0.110 M = 2.48 mmol Ca(OH)2 = 4.96 mmol OH^-1
2HClO4 + Ca(OH)2 --> 2H2O + Ca(ClO4)2
4.96 mmol HClO4 / 13.4 mL = 0.370 M HClO4

2. KHP molecular weight = 204.23 g/mole
0.924 g / 204.23 g/mole = 0.00452 mole
Ba(OH)2 + 2KHP --> 2H2O + Ba(KP)2
0.00452 / 2 = 0.00226 moles of Ba(OH)2 will neutralize 0.00452 moles KHP
0.0254 L x M = 0.00226 mole
Ca(OH)2 = 0.088M

28.1mL x 0.088 M = 2.47 mmol Ca(OH)2 = 4.94 mmol OH^-1
2HClO4 + Ba(OH)2 --> 2H2O + Ba(ClO4)2
4.94 mmol HClO4 / 26 mL = 0.19 M HClO4

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