Question

A 0.533 g sample of a carbonate salt is added to a flask. The only thing you know is that per formula unit, there is only one carbonate anion, but the identity of the cation (or cations) is unknown. To determine the identity of the cations, 15.0 mL of 1.00 M sulfuric acid is added to the flask. After the resulting reaction completes, excess acid remains. The remaining solution is titrated with 0.5905 M NaOH(aq) and reaches the equivalence point when 29.393 mL of base is added. What is the identity of the compound?

Answer 1

`\mathbf{CrCO_3}`

Step-by-step explanation:

The total mole of
`H_2SO_4` added = 15 × 1 × 10⁻³

= 15 × 10⁻³ mole

= 15 mmoles

the number of moles of NaOH added in order to neutralize the excess acid = 0.5905 × 29.393 = 17.36 mmoles

the equation for the reaction is:

2NaOH +
`H_2SO_4` -------->
`Na_2SO_4 +2H_2O`

i.e

2 moles of NaOH react with
`H_2SO_4`

1 moles of NaOH will react with 1/2
`H_2SO_4`

17.36 mmoles of NaOH = 1/2 × 17.36 mmoles of
`H_2SO_4`

= 8.68 mmoles

Number of moles of
`H_2SO_4` that react with MCO₃ = Total moles of
`H_2SO_4` added - moles of
`H_2SO_4` reacted with NaOH

= (15 - 8.68) mmoles

= 6.32 mmoles

`H_2SO_4` +
`MCO_3` ------>
`M_2SO_4` +
`H_2O` +
`CO_2`

1 mole of
`H_2SO_4` react with 1 mole of
`MCO_3`

6.32 mmoles of
`H_2SO_4` = 6.32 mmoles of
`MCO_3`

number of moles of
`MCO_3` = 6.32 10⁻³ moles

mass of
`MCO_3` (carbonate salt) = 0.533 g

molar mass of
`MCO_3` = (M+60)g/mol

We all know that ;

number of moles = mass/molar mass

Then:

6.32 10⁻³ = 0.533 / (M+60)

(M+60) = 0.533/ 6.32 10⁻³

M + 60 = 84.34

M = 84.34 - 60

M = 24.34

Thus the element with the atomic mass of 24.34 is Chromium

The chemical formula for the compound is :
`\mathbf{CrCO_3}`