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ENV-32

Total hardness as CaCO3

TH = 2.497 [Ca] + 4.118 [Mg] in mg/L as CaCO3.

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Water chemistryAPHA 2340

Governing equation

TH=2.497[Ca]+4.118[Mg]\mathrm{TH}=2.497[\mathrm{Ca}]+4.118[\mathrm{Mg}]

[Ca]∈[0,200][Ca] \in \left[0, 200\right]

where

[Ca]
Calcium (mg/L)
[Mg]
Magnesium (mg/L)
TH
Total hardness (mg/L as CaCO3)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

TH=2.497[Ca]+4.118[Mg]\mathrm{TH}=2.497[\mathrm{Ca}]+4.118[\mathrm{Mg}]

[Ca]∈[0,200][Ca] \in \left[0, 200\right]

[Ca]=0[Ca] = 0TH=49.42TH = 49.42

②

Domain of the variable — blue

[Ca]min⁡=0,[Ca]max⁡=200[Ca]_{\min} = 0,\quad [Ca]_{\max} = 200

[Ca]∈[0,200][Ca] \in \left[0, 200\right]

[Ca]=0[Ca] = 0[Ca]=0[Ca] = 0

③

Response curve — green

TH=2.497[Ca]+4.118[Mg]\mathrm{TH}=2.497[\mathrm{Ca}]+4.118[\mathrm{Mg}]

[Ca]∈[0,200][Ca] \in \left[0, 200\right]

[Ca]=0[Ca] = 0TH=49.42TH = 49.42

④

Rate of change — amber

dTHd[Ca]\frac{\mathrm{d}TH}{\mathrm{d}[Ca]}

[Ca]∈[0,200][Ca] \in \left[0, 200\right]

[Ca]=1.258[Ca] = 1.258TH′=2.497TH^{\prime} = 2.497

Lecture brief

Historical brief

Streeter–Phelps (1925) oxygen sag, settling theory and Guldberg–Waage kinetics made water and air quality a rate problem. The lab computes sag, overflow and a snapshot of reactor mass balance. This sheet (ENV-32 — Total hardness as CaCO3) is the form associated with APHA 2340. Working symbols: [Ca][Ca], [Mg][Mg] →\rightarrow THTH. Hardness is the equivalent CaCO3 of divalent cations. The factors convert mg/L Ca and Mg to mg/L CaCO3.

Purpose

Purpose: compute THTH from [Ca][Ca], [Mg][Mg] in Environmental via TH=2.497[Ca]+4.118[Mg]\mathrm{TH}=2.497[\mathrm{Ca}]+4.118[\mathrm{Mg}] TH = 2.497 [Ca] + 4.118 [Mg] in mg/L as CaCO3. Use it when a real environmental question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given [Ca]=40.000 mg/L[Ca] = 40.000\,\mathrm{mg/L}, [Mg]=12.000 mg/L[Mg] = 12.000\,\mathrm{mg/L}, the governing relation TH=2.497[Ca]+4.118[Mg]\mathrm{TH}=2.497[\mathrm{Ca}]+4.118[\mathrm{Mg}] yields TH=149.3 mg/LasCaCO3TH = 149.3\,\mathrm{mg/L as CaCO3}. A beaker, Ca and Mg bars, a hardness scale. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Outputs

  • Total hardness TH149.3 mg/L as CaCO3
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Narration of this film

A beaker, Ca and Mg bars, a hardness scale.

Hardness is the equivalent CaCO3 of divalent cations. The factors convert mg/L Ca and Mg to mg/L CaCO3.

Reading speed

Watch on YouTube