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?

Expert's answer

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,


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.


 Initial milimoles of HCl

= "C \\times V"

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

therefore, milimoles of HCl reacted with CaCO3 are,

=(initial -back titrated)


"=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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