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A student determined the $\ce{Ca^{2+}}$ ion content in a sample of nonfat dry milk. For complete titration of the $\ce{Ca^{2+}}$ ion present, a sample weighting $1.483 ~\mathrm{g}$ required $41.33 ~\mathrm{mL}$ of $1.183 \times 10^{-2} ~\mathrm{M}$ EDTA solution. Calculate the mass of the $\ce{Ca^{2+}}$ ions in the titrated sample.

So I did:

$41.33 ~\mathrm{mL} = 0.04133 \mathrm{L}$

$(1.183 \times 10^{-2}~\mathrm{mol\, L^{-1}})\times (0.04133 \mathrm{L}) = 4.89 \times 10^{-4} ~\mathrm{mol}\, \ce{Ca^{2+}}$

$(4.89 \times 10^{-4} ~\mathrm{mol}\, \ce{Ca^{2+}}) \times 40.078 ~\mathrm{g\, (mol\, \ce{Ca^{2+}})^{-1}} = 0.0196 ~\mathrm{g}\, \ce{Ca^{2+}}$

My question is:

A $25.06$ oz box of the powdered milk analyzed above costs $5.46$.

Calculate the cost of $\pu{1.00 g}$ of $\ce{Ca^{2+}}$ ion as provided by this brand of dry milk.

How do I do that?

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  • 3
    $\begingroup$ Carrying the units through the calculation is a good decision! Nice example for a showing progress in homework question. $\endgroup$ – Klaus-Dieter Warzecha Feb 21 '14 at 9:15
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First you calculate the value $v$ of the powder per gram: $$ 25.06~\mathrm{oz} = 710.44~\mathrm{g} \; \hat{=} \; 5.46~\$ \Rightarrow v=0.007685~\mathrm{\$\, (g\, Powder)^{-1}}$$

Then you calculate the mass of powder ($m$) that contains one gram of calcium ions: $$ m=\frac{1.483~\mathrm{g\, Powder}}{0.0196~\mathrm{g\, \ce{Ca^{2+}}}} = 75.66~\mathrm{\frac{g\, Powder}{g\, \ce{Ca^{2+}}}} $$

Finally, you multiply the two results for the price $p$: $$ p = v\times m = 0.58~\mathrm{\$\, (g\, \ce{Ca^{2+}})^{-1}}$$

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