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o-na [289]
3 years ago
11

The ksp of copper(ii) carbonate, cuco3, is 1.4 × 10-10. calculate the molar solubility of this compound.

Chemistry
1 answer:
mafiozo [28]3 years ago
3 0
As,
                              CuCO₃    ⇆  Cu²⁺  +  CO₃²⁻
So,
                             Kc  =  [Cu²⁺] [CO₃²⁻] / CuCO₃
Or,
                             Kc (CuCO₃)  =  [Cu²⁺] [CO₃²⁻]
Or,
                             Ksp  =  [Cu²⁺] [CO₃²⁻]
As,
Ksp  = 1.4 × 10⁻¹⁰
So,
                            1.4 × 10⁻¹⁰  =  [x] [x]
Or,
                             x²  =  1.4 × 10⁻¹⁰ 
Or,
                             x  =  1.18 × 10⁻⁵ mol/L
To cahnge ito g/L,
                             x  =  1.18 × 10⁻⁵ mol/L × 123.526 g/mol

                   
         x  =  1.45 × 10⁻³ g/L
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To convert the moles to grams we multiply the moles by the molar mass of the compound. Molar mass of the compound is the sum of atomic masses of all the atoms present in it.

molar mass of  = atomic mass of Hg + 2(atomic mass of I) + 6(atomic mass of O)

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