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A sample of water has its hardness due to only $\ce{CaSO4}$. When this water is passed through on anion exchange resin, $\ce{SO4^2-}$, the $\ce{SO}$ ions are replaced by $\ce{OH-}$. A 25.0 mL sample of water so treated requires 21.58 mL of 0.001 M sulfuric acid for its titration. What is the hardness of water expressed in terms of $\ce{CaCO3}$ in ppm? Assume the density of water is 1.0 g/mL.

I do not understand where does $\ce{CaCO3}$ come from if we started with $\ce{CaSO4}$at the beginning.

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The problem isn't going through gyrations just to drive you crazy. It is really the way that water analysis works. The "overall" hardness is expressed in terms of the "equivalent ppm of $\ce{CaCO3}$".

The two biggest contributors to water hardness are typically $\ce{CaCO3}$ and $\ce{MgCO3}$. Now let's say that you want to buy a water softener to replace cations $\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$ with $\ce{Na^+}$ and the $\ce{CO3^{2-}}$ with $\ce{Cl^-}$. Figuring how many gallons of water you'd use in a week and knowing your water's hardness as "equivalent ppm of $\ce{CaCO3}$" you could figure out how much ion exchange resin you'd need. So there is a functional reason that "equivalent ppm of $\ce{CaCO3}$" is helpful.

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  • $\begingroup$ But I am interested is calcium carbonate present in sample above or I just have to calculate possible amount of calcium carbonate that can be formed from amount of calcium from calcium phosphate. $\endgroup$ – chemistry Feb 19 '17 at 20:21
  • $\begingroup$ You're calculating as if the reaction were $\ce{CaCO3 + H2SO4 -> CaSO4 + H2O + CO2}$ $\endgroup$ – MaxW Feb 19 '17 at 21:23
  • $\begingroup$ I calculated molarity of sulphuric acid and looking at reaction Ca(OH)<sub>2</sub> + H<sub>2</sub>SO<sub>4</sub> -----> CaSO<sub>4</ sub> + H< sub>2 </sub>O AND then found moles of calcium phosphate as it is equal to moles of sulphuric acid. Moles of calcium carbonate is equal to moles of calcium phosphate soi found its mass and found hardness in ppm . I wonder If It is ok. $\endgroup$ – chemistry Feb 19 '17 at 22:21
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    $\begingroup$ @chemistry - Yes, moles of calcium carbonate is equal to moles of calcium phosphate. But did you find the mass of calcium carbonate or that of calcium phosphate to convert to ppm? $\endgroup$ – MaxW Feb 19 '17 at 22:33
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    $\begingroup$ I found mass of calcium carbonate and divaded it by mass of water. $\endgroup$ – chemistry Feb 19 '17 at 22:37
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Reporting hardness in terms of "mg/L as $\ce{CaCO_3}$" is common in water treatment practice. The whole point is to get all of your compounds' concentrations in relatable terms so that you can subtract/add them easily (similar to normality). This is called equivalence. In the case of hardness (or titrations), you are concerned with the equivalent amount of protons a compound can give or release. In the case of calcium sulfate, you have two equivalents because the calcium and sulfate both have a charge of 2 (they give/take two protons). The first thing you need to do is convert calcium sulfate into an equivalent weight (EW), which is the molecular weight divided by the number of equivalents, so $\frac{136.14 g/mol}{2 eq/mole}$ = 68.07 g/eq. Note: the English of your question was a bit difficult to understand; read the problem carefully as you may need the EW of calcium hydroxide, not sulfate. The problem approach is the same though, only the number would change.

Now, to get this in terms of mg/L as $\ce{CaCO_3}$ you use the following equation: (mg/L as $\ce{CaCO_3}$)=(mg/L of the species)($\frac{EW_{CaCO_3}}{EW_{species}}$) Similar to calcium sulfate, calcium carbonate has an equivalence of 2 (+2/-2 charge), so the EW is (100/2=50). The problem you are working on is very simple because you only need to convert one compound - often we need to convert many compounds to find, e.g., the alkalinity of a water.

Hint: To get mg/L species you use the titration information - recall that the equivalent definition we are using refers to the amount of protons given/received, so the titration info tells you just that!

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