Question

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

Answer 1

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`.