A new method to distinguish the hyponatremia of electrolyte loss from that due to pure solvent changes

Abstract

Estimates of solute and solvent changes during electrolyte abnormalities are valid only when either total body water (TBW) or solute content do not change, while it cannot be established which one of these is altered. The present paper provides a method capable of distinguishing these two different conditions. When only solvent changes, the respective concentration ratios of plasma (P) solutes PCl/PNa, Poan/PNa, PCl/Poan (Poan = anions other than Cl) remain unchanged. Moreover, PNa₁/PNa₀ (the ratio of PNa during the derangement over the normal value, indicated by subfix ₁ and ₀, respectively) = PCl₁/PCl₀ = Poan ₁/Poan _0. When these constraints are met, the abnormality is due only to a TBW change, which is easily calculated and corrected. When they are not met, the exact change in Na content is correctly calculated assuming no variation in TBW. These calculations could still be useful even in the presence of TBW modifications, where they represent minimum estimates of electrolyte losses. The formulas were validated by computer simulations generating true electrolyte concentrations, which were then used to back calculate the changes in their contents and extra/intra-cellular volumes. Since the predicted results were significantly correlated with the true data, the method was transferred to 24 patients with electrolyte disturbances, who met the above constraints. The calculated volume changes were significantly correlated with those obtained by body weight measurements (regression coefficient = 0.94, P < 0.0001), while the quantitative estimates of Na deficit predicted the PNa values measured after corrective treatment (P < 0.0001). This new method may prove valuable in diagnosing and treating electrolyte derangements.

Appendices

Appendix I

Flow chart describing the steps of computer simulation and of calculations of Na and water surfeits or deficits, omitting OAN for the sake of simplicity.

Example I: Looking at Fig. 1a, we have:

A: Simulation

• 1. The total content in Osmoles is 2 · PNa₀ · TBW₀ = 280 · 40 = 11,200 mOsm;

• 2. Adding 4.8 L of water (Fig. 1b) changes the osmolality to: Posm₁ = total Osmoles/TBW = 11,200/44.8 = 250 mOsm/kg;

• 3. ECV₁ = 2 · PNa₀ · ECV₀/Posm₁ = 16.8 l;

• 4. PNa₁ = PNa₀ · ECV₀/ECV₁ = 140 · 15/16.8 = 125 mEq/l;

• 5. PCl₁ = PCl₀ · ECV₀/ECV₁ = 105 · 15/16.8 = 93.75;

• 6. PCl₁/PNa₁ = 93.75/125 = 0.75 = PCl₀/PNa₀; PNa₁/PNa₀ = 125/140 = 0.89 = PCl₁/PCl₀;

B: Calculation using only the assumed normal values of PNa₀ and TBW₀, and the measured PNa₁. Since the plasma solute ratios are unchanged, we calculate the pure water change:

• 1. ΔH₂O = (PNa₀ − PNa₁) · TBW₀/PNa₁ = (140 − 125) · 40/125 = + 4.8 l;

Example II: Looking at Fig. 1c, we have:

A: Simulation

• 2. Subtracting 4.8 l of water from the normal content of Fig. 1a changes the osmolality to: Posm₁ = total Osmoles/TBW = 11,200/35.2 = 318.2 mOsm/kg;

• 3. ECV₁ = 2 · PNa₀ · ECV₀/Posm₁ = 13.2 l;

• 4. PNa₁ = PNa₀ · ECV₀/ECV₁ = 140 · 15/13.2 = 159 mEq/l;

• 5. PCl₁ = PCl₀ · ECV₀/ECV₁ = 105 · 15/13.2 = 119.3;

• 6. PCl₁/PNa₁ = 119.3/159 = 0.75 = PCl₀/PNa₀; PNa₁/PNa₀ = 159/140 = 1.136 = PCl₁/PCl₀;

B: Calculation using only the assumed normal values of PNa₀ and TBW₀, and the measured PNa₁. Since the plasma solute ratios are unchanged, we calculate the pure water change:

• 1. ΔH₂O = (PNa₀ − PNa₁) · TBW₀/PNa₁ = (140 − 159) · 40/159 = −4.8 l;

Example III: Looking at Fig. 2a, we have:

A: Simulation

• 1. The total content in Osmoles (Fig. 2a) is 2 · PNa₀ · TBW₀ = 280 · 40 = 11,200 mOsm;

• 2. Subtracting 600 mEq of Na (Fig. 1b) changes the osmolality to:

  $$ {\text{Posm}}_{1} = {\text{total Osmoles}}/{\text{TBW}} = (11200 - 2{\text{ }}\cdot {\text{ }}600)/40 = 250{\text{ mOsm}}/{\text{kg}}; $$

• 3. ECV₁ = (2 · PNa₀ · ECV₀ − 2 · ΔNa)/Posm₁ = (2 · 140 · 15 − 2 · 600)/250 = 12 L;

• 4. PNa₁ = (PNa₀ · ECV₀ − ΔNa)/ECV₁ = (140 · 15 − 600)/12 = 125 mEq/L;

• 5. PCl₁ = (PCl₀ · ECV₀ − ΔCl)/ECV₁ = (105 · 15 − 500)/12 = 89.6;

• 6. PCl₁/PNa₁ = 89.6/125 = 0.717 ≠ PCl₀/PNa₀; PNa₁/PNa₀ = 125/140 = 0.89 ≠ PCl₁/PCl₀ = 0.85;

B: Calculation using only the assumed normal values of PNa₀ and TBW₀, and the measured PNa₁. Since the plasma solute ratios are importantly modified, we calculate the pure solute change:

• 1. ΔNa = (PNa₁ − PNa₀) · TBW₀ = (125 − 140) · 40 = − 600 mEq;

Omitting for simplicity the example from Figure 2c, we can proceed with:

Example IV: We withdraw 600 mEq of Na, 600 mEq of Cl, and 1 l of water from Figure 1a.

A: Simulation

• 1. The total content in Osmoles is 2 · PNa₀ · TBW₀ = 280 · 40 = 11,200 mOsm;

• 2. Subtracting 600 mEq of Na (Figure 1b) changes the osmolality to:

  $$ {\text{Posm}}_{1} = {\text{total Osmoles}}/{\text{TBW }} = {\text{ }}(11200 - 2{\text{ }}\cdot {\text{ }}600)/39 = 256.4{\text{ mOsm}}/{\text{kg}}; $$

• 3. ECV₁ = (2 · PNa₀ · ECV₀ − 2 · ΔNa)/Posm₁ = (4,200 − 2 · 600)/256.4 = 11.7 l;

• 4. PNa₁ = (PNa₀ · ECV₀ − ΔNa)/ECV₁ = (140 · 15 − 600)/11.7 = 128.2 mEq/l;

• 5. PCl₁ = (PCl₀ · ECV₀ − ΔCl)/ECV₁ = (105 · 15 − 600)/11.7 = 83.3;

• 6. PCl₁/PNa₁ = 83.3/128.2 = 0.65 ≠ PCl₀/PNa₀; PNa₁/PNa₀ = 128.2/140 = 0.916 ≠ PCl₁/PCl₀ = 0.793;

B: Calculation using only the assumed normal values of PNa₀ and TBW₀, and the measured PNa₁. As the plasma solute ratios are importantly modified, we calculate the pure solute and ECV change:

• 1. ΔNa = (PNa₁ − PNa₀) · TBW₀ = (128.3 − 140) · 40 = − 468 mEq;

• 2. ECV₁ = (2 · PNa₀ · ECV₀− 2 · ΔNa)/Posm₁ = (4,200 − 2 · 468)/256.4 = 12.7 l; in this example, both the Na deficit and the fall in ECV are underestimated by 132 mEq and 1 l, respectively.

Appendix II

Flow chart for computing data of patients reported in Table 1.

Example I.

A: Calculation of water surfeit in patient number 6, whose PCl/PNa ratio indicated pure volume excess. Since we do not know TBW₀, instead of Eq. 6 we apply the equivalent equation 6bis:

• 1. ΔH₂O = (PNa₀ − PNa₁) · BW₁ · 0.6/PNa₀ = (140 − 121) · 60 · 0.6/140 = + 4.9 L;

• 2. True ΔH₂O = BW1 − BW₂ = 60.0 − 55.0 = 5 L;

B: Calculation of predicted PNa₂ at the end of correction of the derangement. The balance study executed during correction registered a net Na balance of −311 mEq. As TBW₀ is unknown:

• 1.
  $$ \begin{aligned}{} {\text{Posm}}_{2} = & {\text{ }}[({\text{Posm}}_{1} \cdot {\text{ TBW}}_{1} + 2{\text{ }}\cdot {\text{ Na balance}})]/({\text{TBW}}_{1} + {\text{ water balance}}); \\ = & [(2{\text{ }}\cdot {\text{ }}121{\text{ }}\cdot {\text{ }}60{\text{ }}\cdot {\text{ }}0.6 - 2{\text{ }}\cdot {\text{ }}311)/(36 - 4.9) = 260{\text{ mOsm}}/{\text{kg}}; \\ \end{aligned} $$

  (11bis)
• 2.
  $$ {\text{ECV}}_{{\text{2}}} {\text{ = (ECV}}_{{\text{1}}} {\text{ }} \cdot {\text{ Posm}}_{{\text{1}}} {\text{ + 2 }} \cdot {\text{ Na balance)/Posm}}_{{\text{2}}}; {\text{ (12bis)}} $$

  since ECV₀ is unknown, ECV₁ must be calculated firstly:

  • (a) Calculated TBW₀ = BW₁ · 0.6 − ΔH₂O = 60 · 0.6 − 4.9 = 31.1 l;

  • (b) Calculated ECV₀ = calculated TBW₀ · 0.375 = 31.1 · 0.375 = 11.7 l; since PNa₀ · ECV₀ = PNa₁ · ECV₁ in pure volume loss:

  • (c) ECV₁ = PNa₀ · calculated ECV₀/PNa₁ = 140 · 11.7/121 = 13.5 L; = BW₁ · 0.6 · 0.375 = 60 · 0.6 · 0.375 = 13.5 l (direct calculation);

  • (d) ECV₂ = (13.5 · 2 · 121 − 2 · 311)/260 = 10.1;

• 3.
  $$ \begin{aligned}{} {\text{PNa}}_{2} = & ({\text{PNa}}_{1} \cdot {\text{ ECV}}_{1} - \Delta {\text{Na}})/{\text{ECV}}_{2} ; \\ = & (121{\text{ }}\cdot {\text{ }}13.5 - 311)/10.1 = 131{\text{ mEq}}/{\text{l}}; \\ \end{aligned} $$

• 4. Since the correction was interrupted before reaching the normal PNa, the volume change must be estimated as follows:

  • (a)
    $$ \begin{aligned}{} {\text{Estimated }}\Delta {\text{Na}} = & ({\text{PNa}}_{2} \cdot {\text{ ECV}}_{2} ) - ({\text{PNa}}_{1} \cdot {\text{ ECV}}_{1} ); \\ = & (132\cdot 10.1) - (121\cdot 13.5) = - 300{\text{ mEq}}; \\ \end{aligned} $$

    (14)
  • (b)
    $$ \begin{aligned}{} {\text{Estimated TBW}}_{2} = & [({\text{PNa}}_{1} \cdot {\text{ TBW}}_{1} ) + {\text{Estimated }}\Delta {\text{Na}}]/{\text{PNa}}_{2} ; \\ = & [(121{\text{ }}\cdot {\text{ }}36) - 300]/131 = 31.0{\text{ l}}; \\ \end{aligned} $$

    (15)
  • (c)
    $$ \begin{aligned}{} {\text{Calculated }}\Delta {\text{TBW}} = & {\text{TBW}}_{1} - {\text{Estimated TBW}}_{2} ; \\ = & 36{\text{ }} - {\text{ }}31 = 5.0{\text{ l}}. \\ \end{aligned} $$

    (16)

Example II: Calculation of pure Na loss in patient 13, whose PCl/PNa indicated pure Na deficit. We compute:

• 1. ΔNa = (PNa₁  − PNa₀) · BW₁ · 0.6 = (108 − 140) · 80 · 0.6 = − 1,536 mEq; even though the preceding calculation assumes a “pure” Na loss, our formulas allow us to compute an associated volume change;

• 2. The volume change computation requires using (11bis)–(16)

  • (a)
    $$ \begin{aligned}{} {\text{Posm}}_{2} = & [({\text{Posm}}_{1} \cdot {\text{TBW}}_{1} + 2{\text{ }}\cdot {\text{ Na balance}})]/({\text{TBW}}_{1} + {\text{water balance}}); \\ = & [(2{\text{ }}\cdot {\text{ }}108{\text{ }}\cdot {\text{ }}80{\text{ }}\cdot {\text{ }}0.6{\text{ }} + {\text{ }}2{\text{ }}\cdot {\text{ }}1100)]/({\text{ }}80{\text{ }}\cdot {\text{ }}0.6 - 2){\text{ }} = {\text{ }}273{\text{ mOsm}}/{\text{kg}}; \\ \end{aligned} $$

    (11bis)
  • (b)
    $$ \begin{aligned}{} {\text{ECV}}_{2} = & ({\text{ECV}}_{1} {\text{ }}\cdot {\text{ Posm}}_{1} + 2{\text{ }}\cdot {\text{ Na balance}})/{\text{Posm}}_{2} ; \\ = & (80{\text{ }}\cdot {\text{ }}0.6{\text{ }}\cdot {\text{ }}0.375{\text{ }}\cdot {\text{ }}216{\text{ }} + {\text{ }}2200)/273{\text{ }} = {\text{ }}22.3{\text{ l}}; \\ \end{aligned} $$

    (12bis)
  • (c)
    $$ \begin{aligned}{} {\text{PNa}}_{2} = & ({\text{PNa}}_{1} \cdot {\text{ ECV}}_{1} + {\text{Na balance}})/{\text{ECV}}_{2} ; \\ = & (108{\text{ }}\cdot {\text{ }}80{\text{ }}\cdot {\text{ }}0.6{\text{ }}\cdot {\text{ }}0.375{\text{ }} + {\text{ }}1100)/22.3{\text{ }} = {\text{ }}136.6; \\ \end{aligned} $$

    (13)
  • (d)
    $$ \begin{aligned}{} {\text{Estimated }}\Delta {\text{Na }} = & ({\text{PNa}}_{2} {\text{ }}\cdot {\text{ ECV}}_{2} ) - ({\text{PNa}}_{1} \cdot {\text{ ECV}}_{1} ); \\ = & ({\text{ }}135{\text{ }}\cdot {\text{ }}22.3){\text{ }} - {\text{ }}(108{\text{ }}\cdot {\text{ }}18){\text{ }} = {\text{ }}1066{\text{ mEq}}; \\ \end{aligned} $$

    (14)
  • (e)
    $$ \begin{aligned}{} {\text{Estimated TBW}}_{2} = & [({\text{PNa}}_{1} \cdot {\text{ TBW}}_{1} ) + {\text{Estimated }}\Delta {\text{Na}}]/{\text{PNa}}_{2} ; \\ = & [(108{\text{ }}\cdot {\text{ }}48){\text{ }} + {\text{ }}1066]/135{\text{ }} = {\text{ }}46.3{\text{ l}}; \\ \end{aligned} $$

    (15)
  • (f)
    $$ \begin{aligned}{} {\text{Calculated }}\Delta {\text{TBW}} = & {\text{TBW}}_{1} - {\text{Estimated TBW}}_{2} ; \\ = & 48 - 46.3 = 1.7{\text{ l}}. \\ \end{aligned} $$

    (16)

An appropriate software is available to ease these calculations (SIAE registration 8/16/06, number 0604311, informations from “sonbartoli@libero.it”, quoting reference program number 0601).