Introduction & Context

Ion-exchange columns are widely used in water-softening, demineralisation, hydrometallurgy and pharmaceutical polishing. The selectivity coefficient \(K'_{\text{Ca/Na}}\) quantifies how strongly the resin prefers Ca²⁺ over Na⁺ under a given set of conditions. Knowing the equilibrium partitioning allows engineers to predict resin loading, breakthrough times, regeneration frequency and column size. The calculation below is valid for the trace-calcium regime (low equivalent fraction of Ca²⁺ in both liquid and solid phases) and is routinely embedded in process simulators and control software.

Methodology & Formulas

Convert solution concentrations to equivalent fractions
\[ u_i = \frac{N_i}{\sum_j N_j} \] where \(N_i\) is the normality of ion \(i\) (eq L^-1).
Trace-calcium equilibrium (simplified separation factor)
When \(u_{\text{Ca}} \ll 1\) and \(v_{\text{Ca}} \ll 1\) the resin-phase equivalent fraction of calcium is \[ v_{\text{Ca}} = K'_{\text{Ca/Na}}\; \frac{u_{\text{Ca}}^{\,z_{\text{Na}}}}{u_{\text{Na}}^{\,z_{\text{Ca}}}} \] with \(z_{\text{Na}}=1\) and \(z_{\text{Ca}}=2\). Sodium fraction on the resin follows by closure: \[ v_{\text{Na}} = 1 - v_{\text{Ca}} \]
Physical bounds
Clamp the calculated value to the interval [0,1]: \[ v_{\text{Ca}} = \max\!\bigl(0,\; \min(1,\; v_{\text{Ca}})\bigr) \]

Regime of Validity
Parameter	Limit	Remarks
Total normality	\(\leq 0.1\) N	Correlation derived for dilute solutions
Equivalent fraction Ca²⁺ in liquid	\(\ll 1\)	Trace approximation invoked
Resin loading Ca²⁺	\(\ll 1\)	Linearised isotherm

All symbols are dimensionless unless units are explicitly shown. Temperature and pressure are assumed constant at 25 °C and 1 bar; \(K'\) is treated as invariant over the stated range.

The selectivity coefficient (often written as K_A/B) is a dimensionless number that tells you how much the resin prefers ion A over ion B when both are present. A value >1 means the resin grabs A more strongly; <1 means it prefers B. In design, it lets you predict:

Leakage of the less-preferred ion in the effluent
Regenerant dosage needed to reach water-quality targets
Whether a cheaper, lower-capacity resin can still meet spec because it “likes” the contaminant ion better

Ignoring it leads to under- or over-designed columns and blown OPEX budgets.

Resin vendors publish tables for their products—always check the exact resin name and ionic form
EPA Methods 833 and ASTM D5601 give lab protocols if you need to measure your own
For mixed brines, run a bench-scale column test at site temperature and ionic strength; coefficients shift with TDS and temperature
Keep a plant logbook—coefficients drift after 500–1000 cycles due to resin aging

Yes—higher temperature generally lowers the coefficient (resin becomes less picky). A 10 °C rise can drop K by 5–15 %. Correct via the van’t Hoff form: K_T2 = K_T1 exp[ΔH/R (1/T1 – 1/T2)]. If your winter/summer feed varies by >15 °C, size the column for the worst-case (lowest K) or add a tempering loop.

Only if you convert to a corrected selectivity coefficient that accounts for charge. Use the Eisenman equation or the Gaines-Thomas convention:

K_Na/Ca = K_apparent × (C_Na/C_Ca)^½
Always work in equivalent ionic fractions, not mass concentrations, or your isotherm will be curved and misleading

Worked Example – Softening a Brackish Cooling-Tower Make-up Water

A small power plant needs to soften 50 m³ h^-1 of make-up water at 25 °C and 1 bar. A strong-acid cation resin in the Na-form is available; the goal is to predict the equilibrium loading of Ca²⁺ and Na⁺ when the column is first brought on-line.

Knowns

Resin capacity: 2.0 eq L^-1
Selectivity coefficient (K′): 3.5 (dimensionless)
Ionic charges: Z_Ca = 2, Z_Na = 1
Total normality of inlet water: 0.100 N
Equivalent fractions in solution: u_Ca = 0.010, u_Na = 0.990
Temperature: 25 °C
Pressure: 1 bar

Step-by-Step Calculation

Write the binary exchange reaction and equilibrium expression: \[ \mathrm{Ca^{2+} + 2NaR \rightleftharpoons CaR_2 + 2Na^+} \qquad K' = \frac{v_{\mathrm{Ca}} \cdot u_{\mathrm{Na}}^2}{u_{\mathrm{Ca}} \cdot v_{\mathrm{Na}}^2} \] where \(v_i\) is the equivalent fraction in the resin phase.
Impose the stoichiometric constraints: \[ v_{\mathrm{Ca}} + v_{\mathrm{Na}} = 1 \qquad u_{\mathrm{Ca}} + u_{\mathrm{Na}} = 1 \]
Insert known values into the selectivity expression: \[ 3.5 = \frac{v_{\mathrm{Ca}} \cdot (0.990)^2}{0.010 \cdot (1 - v_{\mathrm{Ca}})^2} \]
Solve the resulting quadratic for \(v_{\mathrm{Ca}}\): \[ v_{\mathrm{Ca}} = 0.036 \qquad v_{\mathrm{Na}} = 1 - 0.036 = 0.964 \]
Convert equivalent fractions to resin loadings (eq L^-1): \[ q_{\mathrm{Ca}} = v_{\mathrm{Ca}} \cdot \text{capacity} = 0.036 \cdot 2.0 = 0.072\ \mathrm{eq\ L^{-1}} \] \[ q_{\mathrm{Na}} = v_{\mathrm{Na}} \cdot \text{capacity} = 0.964 \cdot 2.0 = 1.928\ \mathrm{eq\ L^{-1}} \]

Final Answer

At equilibrium the resin contains 0.072 eq L^-1 of Ca²⁺ and 1.928 eq L^-1 of Na⁺, confirming preferential uptake of divalent calcium despite its low solution fraction.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle