Answer to Question #186759 in General Chemistry for Kyle

Question

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?

Answer 1

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,

$H C l ( a q ) + K O H ( a q ) → K C l ( a q ) + H 2 O ( l )$

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 $[ C a C O_ 3 ] = 100$

mas of $[ Ca C O_ 3 ]$

$= 100 \times0.985 \times 10^{-3} \\=0.0985 g$

now, percentage of calcium carbonate in chalk

$= {0.0985\over 0.125} × 100 = 78.8$ %