1.00 g of alkali metal carbonate, M_2CO_3, is dissolved in water and made up to 250 cm^3 in a volumetric flask. 25.00 cm^3 portions of this solution are titrated against 0.113M hydrochloric acid. The average titre is 12.80cm^3. What element is M?

1.00 g of alkali metal carbonate, M_2CO_3, is dissolved in water and made up to 250 cm^3 in a volumetric flask.
25.00 cm^3 portions of this solution are titrated against 0.113M hydrochloric acid. The average titre is 12.80cm^3
Calculate:
a. The moles of HCl in the titre
b. The moles of M_2CO_3 in 25 cm3 of solution
c. The moles of M_2CO_3 in 250 cm3 of solution
d. The relative formula mass of M_2CO_3
e. The identity of M

1 Answer
Nov 14, 2017

One necessary initial step is to write the stoichiometric equation....

M_2CO_3(s) + 2HCl(aq) rarr2MCl(aq) + H_2O(l) + CO_2(g)uarr

Explanation:

"a. Moles of HCl in titre:"

12.8*cm^3xx10^-3*L*cm^-3xx0.113*mol*L^-1=1.45xx10^-3*mol

"b. Moles of carbonate in 25 mL aliquot:"

(12.8*cm^3xx10^-3*L*cm^-3xx0.113*mol*L^-1)/2=7.23xx10^-4*mol

"c. Moles of carbonate in 250 mL:"

7.23xx10^-4*molxx10=7.23xx10^-3*mol

"d. Formula mass of carbonate:"

=(1.00*g)/(7.23xx10^-3*mol)=138.3*g*mol^-1

"e. Identity of metal and metal carbonate:"

We know that the formula mass of M_2CO_3-=138.3*g*mol^-1.

Since the mass of CO_3^(2-) is 60*g*mol^-1, the mass of the metal is....(138.3-60.0*g*mol^-1)/2=39.1*g*mol^-1.

And a quick glance at the Periodic Table confirms that we gots potassium metal, and the carbonate is K_2CO_3.