Use the Relereaces to access important valwet i neoded for thie goesdien. A stud
ID: 569462 • Letter: U
Question
Use the Relereaces to access important valwet i neoded for thie goesdien. A student is asked to standardize a solution of calcium hydroxide. He weighs out 106 g potassiunm hydrogen phdhalate (KHCalilLO, teat this as a monoprotic acid) It requires 32.9 ml. of cakium hydroxide to reach the endpoint. A. What is the molarity of the calchum bydrozide solution? This calcium hydreside solution is then used to titrate an unknown solution of perchloric acid B. If 104 mL of the calcium bydroside solution is requaired to neutralize 114mL of perchloric acid, what is the melarity of the perchloric acid solution? Subrmit Answer Retry Entire Group 9 more group attempts remaining NedExplanation / Answer
1) Molar mass of potassium hydrogen phthalate, KHC8H4O4 = (1*39.0983 + 5*1.008 + 8*12.01 + 4*15.9994) g/mol = 204.2159 g/mol.
Mole(s) of potassium hydrogen phthalate (KHP) corresponding to 1.06 g KHP = (1.06 g KHP)*( 1 mole KHP/204.2159 g KHP) = 0.00519 mole.
(A) KHP is a monoprotic acid and reacts with calcium hydroxide, Ca(OH)2 as below.
Ca(OH)2 (aq) + 2 KHC8H4O4 (aq) ---------> Ca(KC8H4O4)2 (aq) + 2 H2O (l)
As per the stoichiometry of the reaction,
1 mole Ca(OH)2 = 2 moles KHP.
Therefore, mole(s) of Ca(OH)2 neutralized = (0.00519 mole KHP)*(1 mole Ca(OH)2/2 mole KHP) = 0.002595 mole Ca(OH)2.
Volume of Ca(OH)2 required to reach the equivalence point = 32.9 mL = (32.9 mL)*(1 L/1000 mL) = 0.0329 L.
Molarity of Ca(OH)2 solution = (moles of Ca(OH)2)/(volume of Ca(OH)2 in L) = (0.002595 mole)/(0.0329 L) = 0.07887 mol/L = 0.07887 M (ans).
(B) Write down the balanced chemical equation for the reaction between Ca(OH)2 and perchloric acid, HClO4.
Ca(OH)2 (aq) + 2 HClO4 (aq) -------> Ca(ClO4)2 (aq) + 2 H2O (l)
As per the stoichiometric equation,
1 mole Ca(OH)2 = 2 moles HClO4.
Mole(s) of Ca(OH)2 in 10.4 mL of 0.07887 M Ca(OH)2 = (10.4 mL)*(1 L/1000 mL)*(0.07887 M) = 0.000820 mole.
Mole(s) of HClO4 neutralized = (0.000820 mole Ca(OH)2)*(2 moles HClO4/1 mole Ca(OH)2) = 0.00164 mole.
Molarity of HClO4 = (moles of HClO4 neutralized)/(volume of HClO4 in L) = (0.00164 mole)/[(11.4 mL)*(1 L/1000 mL)] = 0.1438 mol/L = 0.1438 M (ans).
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