A. Calculate the number of millimoles of naoh used to titrate leftover acid B. N

Question

A. Calculate the number of millimoles of naoh used to titrate leftover acid
B. Number of millimoles of left over acid.
C. Number of millimoles of Acid added initially ( 4.00 ml of 3.00M HCl)

Additional information: naoh molarity is 0.1893M HCl used is 4ml of a 3.00M
A. Calculate the number of millimoles of naoh used to titrate leftover acid
B. Number of millimoles of left over acid.
C. Number of millimoles of Acid added initially ( 4.00 ml of 3.00M HCl)

Additional information: naoh molarity is 0.1893M HCl used is 4ml of a 3.00M

B. Number of millimoles of left over acid.
C. Number of millimoles of Acid added initially ( 4.00 ml of 3.00M HCl)

Additional information: naoh molarity is 0.1893M HCl used is 4ml of a 3.00M

Explanation / Answer

(a): Run - 1: Volume of NaOH used = 0.20 mL

Molarity of NaOH = 0.1893M

Hence millimoles of NaOH used = MxV(mL) = 0.1893M x 0.20 mL = 0.03786 millimol (answer)

Run - 2: Volume of NaOH used = 0.16 mL

Molarity of NaOH = 0.1893M

Hence millimoles of NaOH used = MxV(mL) = 0.1893M x 0.16 mL = 0.030288 millimol (answer)

(b): Number of millimoles of NaOH used is equal to the number of millimoles of acid left over.

Run-1: Millimol of acid left over = 0.03786 millimol (answer)

Run-2: Millimol of acid left over = 0.030288 millimol (answer)

(c): Run - 1: Volume of HCl used = 4.00 mL

Molarity of HCl = 3.00 M

Hence millimoles of acid added initially = MxV(mL) = 3.00 x 4.00 mL = 12.00 millimol (answer)

Run - 2: millimoles of acid added initially = MxV(mL) = 3.00 x 4.00 mL = 12.00 millimol (answer)

(d) Run-1: Millimol of HCl that reacted with the tablet = 12.00 millimol - 0.03786 millimol

= 11.96214 millimol (answer)

Run-2: Millimol of HCl that reacted with the tablet = 12.00 millimol -  0.030288 millimol

= 11.969712 millimol (answer)

(e): Run-1: millimoles of CaCO3 in the tablet = 11.96214 millimol / 2 = 5.98107 millomol (answer)

Run-2: millimoles of CaCO3 in the tablet = 11.969712 millimol / 2 = 5.984856 millimol

(f) Run - 1: mass of CaCO3 = 5.98107 millomol x 100.0 g/mol = 598.107 mg

Run - 2: mass of CaCO3 = 5.984856 millomol x 100.0 g/mol = 598.4856 mg

average mass of CaCO3 = (598.107 mg + 598.4856 mg ) / 2 = 598.296 mg

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