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My question is regarding the use of sodium bisulphate to lower the hardness of water. I am attempting to come up with a label to let consumers know how much product (dry acid or sodium bisulphate) is needed for varying hardness values.

Using the equation $$\ce{CaCO3 + 2NaHSO4 -> Na2SO4 + CaSO4 + CO2 + H2O}$$I have found the following ratios:

$\pu{120 mg/L}$ $\ce{CaCO3}$ requires $\pu{0.288g}$ $\ce{NaHSO4}$, $\pu{180 mg/L}$ $\ce{ CaCO3 }$ requires $\pu{0.432g}$ $\ce{NaHSO4}$, etc.

My question: How much calcium will react with bisulfate? And how can I use this information to determine the change in hardness? Will starting pH of the water affect the calculation?

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closed as off-topic by Mithoron, A.K., Todd Minehardt, airhuff, M.A.R. ಠ_ಠ Feb 22 '18 at 17:48

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  • $\begingroup$ Why is this on hold? This is not a homework problem, this is a valid softening question. @Mithoron $\endgroup$ – prof.kvothe Feb 24 '18 at 15:45
  • $\begingroup$ You are in equivalents already with 120 mg/L as CaCO3 - that's your hardness and the amount you need to remove. The EW of SB is MW/2 or 60 g/eq. The equivalent weight of CaCO3 is 50. Thus, for 120 mg/L as CaCO3 of hardness, you will need 120*(60/50) of SB or 144 mg/L. OR a better way to look at it, you need 1.2 times the SB (mg/L) that you have in hardness (mg/L CaCO3). So, if you add 288 mg/L, you should have zero hardness left because it's ~2x the stoichiometric requirement. The pH will drop and hinder solubility, which may be where the 2x comes from. I'll ans. better when unlocked. $\endgroup$ – prof.kvothe Feb 24 '18 at 15:57