solution. 3 imes B. Determination of the molar of a basic oxide. Weigh out accur
ID: 557298 • Letter: S
Question
solution. 3 imes B. Determination of the molar of a basic oxide. Weigh out accurately to the nearest 0.1 mg about 0.1 g of the basic oxide provided and transfer the sample carefully to a 125 mL Erlenmeyer flask. Now carefully add with a pipet or buret exactly 25.00 mL of the standard HCl solution. Swirl the flask gently until all of the oxide has reacted. Once all the oxide has reacted, the solution should be clear. However, it may be necessary to warm the mixture to ensure that the sample has reacted completely. Cool to room temperature if necessary, add 4 drops of phenolphthalein indicator, and titrate with the standard NaOH solution until a faint pink color is obtained. Repeat the determination with another accurately sample of the basic oxide. From your results calculate the equivalent weight of a basic oxide.Explanation / Answer
Trial 1:
Mol of HCl added = 0.2946 mol/L * (25/1000)L = 7.365*10-3mol
Mol of NaOH used = 0.09628 mol/L * (35.6/1000)L = 3.4275*10-3mol
Mol of HCl reacted = Mol of NaOH used at equivalent point so
Mol of HCl reacted = 3.4275*10-3mol
Mol of HCl left = 7.365*10-3mol - 3.4275*10-3mol = 3.9375*10-3 mol
Molar mass of oxide = 0.100 gm / 3.9375*10-3 mol = 25.3968 g/mol
Trial 2:
Mol of HCl added = 0.2946 mol/L * (25/1000)L = 7.365*10-3mol
Mol of NaOH used = 0.09628 mol/L * (35.4/1000)L = 3.4083*10-3 mol
Mol of HCl reacted = 3.4083*10-3 mol
Mol of HCl left = 7.365*10-3 - 3.4083*10-3 = 3.9566*10-3mol
Molar mass = 0.098 gm / 3.9566*10-3mol = 24.7681 gm/mol
Trial 3:
Mol of HCl added = 0.2946 mol/L * (25/1000)L = 7.365*10-3mol
Mol of NaOH used = 0.09628 mol/L * (35.7/1000)L = 3.4371*10-3 mol
Mol of HCl reacted = 3.4371*10-3 mol
Mol of HCl left = 7.365*10-3 - 3.4371*10-3 = 3.9279*10-3 mol
Molar mass of oxide = 0.097gm / 3.9279*10-3 mol = 24.6951 gm/mol
Avg Molar mass = (25.3968 + 24.7681+24.6951)/3 = 24.9533 gm/mol
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