Question

So this is supposed to be the formula but I keep getting it wrong:

"Equivalent millimoles HCl/mg sample = ( (concentration HCl in mmol/mL)(Volume HCl in mL) - (concentration NaOH in mmol/mL)(Volume NaOH in mL) ) / ( (1000 mg/g)(mass antacid sample in grams ) )"

My molarity HCl: 0.366

My molarity NaOH: 0.099

Part IIC Scoring Scheme: 3-3-2-1 Based upon the molarities of the standardized HCl and standardized NaOH, and the volumes of each added to analyze the antacid samples, calculate the MILLIMOLES of HCI needed to neutralize the tablet per MILLIGRAM of sample Use the same sequence as you entered the mass of the samples and report your results to 4 significant figures Entry # mass tablet(g) 3.887 3.887 3.887 mass antacid( 0.6000 0.6000 0.6000 g)Vol HCI(mL) 14.00 14.00 14.00 Vol NaOH(mL)mmoles HCl/mg 5.200 5.000 5.100 #2: #3:

Answer 1

For Entry 1:

Number of millimoles of HCl = Volume of HCl (in mL) * Molarity of HCl = 14.00 mL * 0.366M = 5.124 mmoles

Number of millimoles of NaOH = Volume of NaOH (in mL) * Molarity of NaOH = 5.200 mL * 0.099M = 0.5148 mmoles

Millimoles of HCl left = Number of millimoles of HCl - Number of millimoles of NaOH = (5.124 - 0.5148) mmoles = 4.6092 mmoles

Mass of antacid tablet = 0.6000g = 600 mg

mmoles of HCl/mg = 4.6092 mmoles/600 mg = 0.007682 mmoles/mg

For Entry 2:

Number of millimoles of HCl = Volume of HCl (in mL) * Molarity of HCl = 14.00 mL * 0.366M = 5.124 mmoles

Number of millimoles of NaOH = Volume of NaOH (in mL) * Molarity of NaOH = 5.000 mL * 0.099M = 0.495 mmoles

Millimoles of HCl left = Number of millimoles of HCl - Number of millimoles of NaOH = (5.124 - 0.495) mmoles = 4.629 mmoles

Mass of antacid tablet = 0.6000g = 600 mg

mmoles of HCl/mg = 4.629 mmoles/600 mg = 0.007715 mmoles/mg

For Entry 3:

Number of millimoles of HCl = Volume of HCl (in mL) * Molarity of HCl = 14.00 mL * 0.366M = 5.124 mmoles

Number of millimoles of NaOH = Volume of NaOH (in mL) * Molarity of NaOH = 5.100 mL * 0.099M = 0.5049 mmoles

Millimoles of HCl left = Number of millimoles of HCl - Number of millimoles of NaOH = (5.124 - 0.5049) mmoles = 4.6191 mmoles

Mass of antacid tablet = 0.6000g = 600 mg

mmoles of HCl/mg = 4.6191 mmoles/600 mg = 0.007698 mmoles/mg