Problem Set #3 - Ions and Membrane Potentials - Answer Key
Definitions and Formulas
Concentration Equations:
C=S/V
S=CxV
Nernst Equation:
Eion = 61 * log(Cout/Cin)
z
where: Eion is the equilibrium potential of the ion in millivolts (mV)
z is the charge on the ion (+1 for K+ and Na+, -1 for Cl–)
Cout and Cin are concentrations of the ion in the ECF and ICF.
log is the base 10 logarithm of the number in parentheses
1. The normal ECF concentration of K+ is approximately 4 mM. How many millimoles of K+
would be contained in the entire 15 L of extracellular fluid?
S = C x V = 4 millimoles/L x 15 L
= 60 millimoles of K+
(recall that mM means “millimoles per liter”)
2. a) If you add 30 millimoles of potassium chloride to 200 mL of distilled water, what is the K+
concentration of this solution, in mM?
= 150 mM K+
b) What is the osmolarity of this solution
≈
300 mOsm
150 mM KCl dissociates into two ions to form approximately a 300 mOsm solution.
C = S/V = 30 millimoles /0.2 L
3. Suppose a patient is mistakenly given 600 mL of the above solution instead of isotonic saline
(NaCl) in an intravenous (IV) infusion.
a) How many millimoles of K+ are in 600 mL of this solution? S = C x V = 150 mM x 0.6 L
= 90 millimoles of K+
+
b) Assume that all the K in given to the patient in (a) above
gets evenly distributed throughout the 15 liters of his ECF.
Calculate the new K+ concentration in the ECF.
New total amount of K+ in the ECF = 60 millimoles + 90 millmoles = 150 millmoles
New concentration of K+ in the ECF = 150 millimoles /15 L = 10 mM
How would this affect the resting membrane potential of cardiac muscle and nerve cells
(depolarize or hyperpolarize, relative to the normal resting potential)?
It would depolarize the resting membrane potential (i.e., make it less negative).
4. Given the following (abnormal) concentrations:
ICF
ECF
K+
150 mM
1.5 mM
Na+
50 mM
160 mM
Cl–
15 mM
150 mM
a) Calculate the equilibrium potentials of K+ Na+ , Cl– using the Nernst equation.
Logarithm table
log (0.01) = –2
log (0.1) = –1
log (1) = 0
log (3.2) = 0.5
log (10) = 1
log (100) = 2
EK = (61/1) x log(1.5/150) = 60 x (–2) = –122 mV
ENa = (61/1) x log(160/50) = 60 x (0.5) = +30.5 mV
ECl = (61/–1) x log(150/15) = –60 x (1) = –61 mV
b) In the case above, if the membrane is permeable only
to K+, what would be the predicted membrane potential?
+
equal to EK = –122 mV
–
c) If the membrane is equally permeable to K and Cl ,
and impermeable to all other ions, what would be the
predicted membrane potential?
midway between EK and ECl ≈ –91 mV