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The differential diagnosis of hypernatraemia in children, with ...

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<strong>The</strong> Diagnosis <strong>of</strong> Salt Poison<strong>in</strong>g Lead<strong>in</strong>g to Hypematraemia <strong>in</strong> Children – September 2009<br />

In fact the amount needed will <strong>in</strong>evitably be more than this, firstly because some <strong>of</strong><br />

the adm<strong>in</strong>istered salt will already have been excreted by the time the <strong>in</strong>itial abnormal<br />

measurement is made, and secondly because (if the salt was adm<strong>in</strong>istered orally) it is<br />

almost certa<strong>in</strong> that there will still be unabsorbed salt <strong>in</strong> the GI tract at the time <strong>of</strong> the <strong>in</strong>itial<br />

measurement. <strong>The</strong> 1 and 3 teaspoon values actually assume (a) <strong>in</strong>stant total absorption, and<br />

(b) no excretion <strong>of</strong> any <strong>of</strong> the adm<strong>in</strong>istered salt at the time <strong>of</strong> measurement.<br />

A method <strong>of</strong> calculat<strong>in</strong>g the excess salt load required to raise the serum sodium a given<br />

amount is as follows:<br />

1. Concentration (mmol/l) = Amount (mmol)<br />

Volume (litres)<br />

2. <strong>The</strong> presumed pre-illness total body amount <strong>of</strong> Na + <strong>in</strong> the child’s body is given by:<br />

140 = Amount <strong>of</strong> Na+ (pre-illness)<br />

Volume (0.65 x wt <strong>in</strong> kilogram)<br />

. . . total body Na + (pre-illness) = 140 x (0.65 x wt)<br />

3. <strong>The</strong> total amount <strong>of</strong> Na + <strong>in</strong> the child’s body at presentation is similarly given by:<br />

total body Na + (at presentation) = Pres [Na + ] x (0.65 x wt)<br />

4. <strong>The</strong> excess total body Na + <strong>in</strong> mmol is given by<br />

Amount (at presentation) – Amount (pre-illness)<br />

This is divided by 17.1 to give the excess Na + <strong>in</strong> grams (17.1 mmol = 1 gram)<br />

Example<br />

A child presents <strong>with</strong> a serum [Na + ] <strong>of</strong> 185mmol/l and a recent wt <strong>of</strong> 15 kilogram. <strong>The</strong><br />

child now weighs 14.6 kilogram<br />

1. Previous total body Na + = 140 x (0.65 x 15) = 1,365 mmol<br />

2. Present<strong>in</strong>g total body Na + = 185 x (0.65 x 14.6) = 1,756 mmol<br />

3. Excess Na + = 390mmol = 22.8 grams<br />

Evidence statement<br />

When calculat<strong>in</strong>g the amount <strong>of</strong> sodium required to raise the serum sodium by a<br />

specified amount, the total body water (approximately 65% <strong>of</strong> the child’s weight)<br />

should be used to determ<strong>in</strong>e the volume <strong>of</strong> distribution. [Grade D]<br />

5.3 How well known among the public is the lethal dose <strong>of</strong> salt?<br />

<strong>The</strong>re is no published <strong>in</strong>formation on this question. Accord<strong>in</strong>gly, the RCPCH <strong>in</strong>cluded a question<br />

about this <strong>in</strong> a survey <strong>of</strong> parents undertaken on behalf <strong>of</strong> the College by BMRB Omnibus.<br />

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