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The calculated value of 26.5 g/L is now used as a factor to determine the
volume of solution that can be prepared. However, the reciprocal of the value must be
used as a factor. If there are 26.5 grams per liter then for every liter there are 26.5
grams of solute. This relationship is true of all factors we have used.
L
106.0 g X  = 4.00 L
26.5 g
Section III. MILLIEQUIVALENT PER LITER (mEq/L) SOLUTIONS
47.
INTRODUCTION
Milliequivalent per liter problems are based on the same principle as normality. If
we examine the definition of normality, we will find that a normal solution contains one
gram equivalent weight per liter of solution. Equivalently, we can say that a normal
solution is one equivalent per liter of solution (Eq/L). A milliequivalent per liter solution
would, therefore, contain one milliequivalent weight per liter of solution.
48.
DISCUSSION
a. A milliequivalent = 0.001 equivalent and 1 equivalent = 1000 milliequivalents.
b. A milliequivalent has the numerical value of one equivalent weight expressed
in mg/mEq.
c. To convert Eq/L to mEq/L and vice versa multiply the concentration by the
conversion factor 1000 mEq/1 Eq as appropriate.
49.
SOLVING MILLIEQUIVALENT PER LITER PROBLEMS
a. Example 1. How much NaCl is needed to prepare 500 mL of a 25.0 mEq/L
NaCl solution?
Solution. Read the problem carefully and determine the desired quantity.
Milligrams of NaCl.
Determine the milliequivalent weight of the compound.
NaCl
Na
23.0
C
+ 35.5
58.5 mg/mmol
MD0837
48



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