# Answer to Question #186759 in General Chemistry for Kyle

Question #186759

The mass of the calcium carbonate in 0.125 g stick of chalk is determined by dissolving the chalk in 50.0 mL of 0.20 M HCl and then titrating any excess HCl with a standardized solution of KOH. A volume equal to 32.12 mL of 0.250 M KOH is required to reach the endpoint. Write the complete computation.

A. How many millimoles of the excess HCl was back-titrated?

B. How many millimoles of HCl actually reacted to the CaCO3 in the chalk sample?

C. What is the percent by mass of calcium carbonate in the chalk?

1
2021-05-06T07:37:30-0400

Calcium carbonate (CaCO3): it is a common componant of limstone in the form of the mineral calcite (i.e, its ionic salt);This compound dissociates into its constituent ions - calcium and carbonate ions - upon dissolution in water. Carbonate is a polyprotic weak base, which reacts with 2 moles of a monoprotic acid.

and it reacts with HCl as follows:

"CaCO_{3(aq)} + 2HCl_{(aq)} \\to CaCl_{2(aq)} + H_2CO_{3(aq)}"

A) now, for How many millimoles of the excess HCl was back-titrated?

we use we use equation,

"H\nC\nl\n(\na\nq\n)\n+\nK\nO\nH\n(\na\nq\n)\n\u2192\nK\nC\nl\n(\na\nq\n)\n+\nH\n2\nO\n(\nl\n)"

from given info. moles of KOH used

= "C \\times V"

= "0.250 \\times (32.12 \\times 10^ {-3})"

="8.03 \\times 10^{-3}"

as in equation,

Since they react in a 1:1 ratio the no. of moles of HCl

must be the same:

n(HCl) back titrated = 8.03 milimoles.

B)

Initial milimoles of HCl

= "C \\times V"

"=50 \\times 0.20 \\\\=10"milimoles

therefore, milimoles of HCl reacted with CaCO3 are,

=(initial -back titrated)

=(10-8.03)

"=1.97" milimoles

C) From the original equation you can see that the no. moles of CaCO3

must be half of this.

"CaCO_{3(aq)} + 2HCl_{(aq)} \\to CaCl_{2(aq)} + H_2CO_{3(aq)}"

therefore milimoles of CaCO3 are

="1.97 \\over2"

="0.985" milimoles

molar mass "[\nC\na\n\nC\nO_\n3\n]\n=\n100"

mas of "[\nCa\nC\nO_\n3\n]"

"=\n100 \\times0.985 \\times 10^{-3} \\\\=0.0985 g"

now, percentage of calcium carbonate in chalk

"=\n{0.0985\\over 0.125}\n\u00d7\n100\n=\n78.8" %

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