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umka2103 [35]
3 years ago
14

A chemist prepares a solution of iron(II) bromide FeBr2 by measuring out 0.55kg of iron(II) bromide into a 500.mL volumetric fla

sk and filling the flask to the mark with water.
Calculate the concentration in /molL of the chemist's iron(II) bromide solution.
Round your answer to 2 significant digits
Chemistry
1 answer:
Sidana [21]3 years ago
7 0
Molar mass of ferrous bromide: 215.65<span>
m(FeBr2): 550/215.65 = 2.55moles
this is dissolved in 500mL of water so in 1L of solution there will be 2.55x2 = 5.1mol/L</span>
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<span>     M(OH)</span>₂<span>(s)        ↔        <span>M</span></span>²⁺<span><span> (aq) +          2OH</span></span>⁻<span><span>(aq)</span></span>

<span>I                                         -                            -</span>

<span>C   -X                                +X                        +2X</span>

<span>E                                        X                          2X</span>

 

<span>Ksp                  =          [M</span>²⁺<span> (aq)] [OH</span>⁻<span>(aq)]</span>²

4.45 * 10∧-12        =          (0.202 + X ) (2X)²

Since X is very small, (0.202 + X ) = 0.202

<span>4.45 * 10<span>-12        </span>=          0.202 * 4X</span>²

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Hence the solubility of <span>M(OH)2 is 2.347 </span>× 10∧-6 M

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