A 68 y/o male diagnosed with congestive heart failure presents to his physician
ID: 3475698 • Letter: A
Question
A 68 y/o male diagnosed with congestive heart failure presents to his physician with distended jugulars and pitting edema of the ankles. His breathing is rapid (20 breaths/min) and pulmonary rales (crackles) are heard bilaterally in the lower lobes of the lungs. He has a pulse rate of 110 beats/min and a BP of 140/88. Blood work shows:
Blood
Values
Urine
Values
Na+ (mEq/L)
128
Na+ (mEq/L)
110
K+ (mEq/L)
3.9
K+ (mEq/L)
80
Mg2+ (mg/dL)
1.7
Mg2+ (mg/day)
19
Ca2+ (mg/dL)
8.9
Ca2+ (mg/day)
105
HCO3 (mEq/L)
30
HCO3
1.7
Creatinine (mg/dl)
1.7
Creatinine (mg/L)
2080
PAH (mg/ml)
0.013
PAH (mg/ml)
5.91
Glucose (mg/dL)
85
Glucose
0
BUN (mg/dL)
14
24hr volume (L)
1.2
pCO2 (mmHg)
45
Osmolarity (mOsm/L)
750
pH
7.31
pH
6.8
(use cellular mechanisms and diagrams)
a) Assuming that a normal male has a plasma osmolality of 290 mOsm and that both a normal male and this Px have a total of 12,000 mOsm in their body fluids, how much extra water does our Px have on board? (5 pts)
b) How would you treat the Px to remove the excess fluid from his body? What would be his final serum osmolality after treatment? Would this alter his blood pressure and why? Ultimately, would this treatment cure the Px? Why? (20pts)
c) Is this Px experiencing an acid-base imbalance? If so, what type and how would the body compensate for this imbalance? (10pts)
Blood
Values
Urine
Values
Na+ (mEq/L)
128
Na+ (mEq/L)
110
K+ (mEq/L)
3.9
K+ (mEq/L)
80
Mg2+ (mg/dL)
1.7
Mg2+ (mg/day)
19
Ca2+ (mg/dL)
8.9
Ca2+ (mg/day)
105
HCO3 (mEq/L)
30
HCO3
1.7
Creatinine (mg/dl)
1.7
Creatinine (mg/L)
2080
PAH (mg/ml)
0.013
PAH (mg/ml)
5.91
Glucose (mg/dL)
85
Glucose
0
BUN (mg/dL)
14
24hr volume (L)
1.2
pCO2 (mmHg)
45
Osmolarity (mOsm/L)
750
pH
7.31
pH
6.8
Explanation / Answer
A. Calculate the osmolality for the patient using the equation
Osmolality=2[Na+conc. In Meq/L] + 0.055[ Glucose Conc.in mg/dl] + 0.36 [ BUN in mg/dl ]
= 2 [128] + 0.055 [85] + 0.36[14]
=265.7mOsm
Total body water of normal patient=12000mOsm
Total body water in litres=
Total body water in mOsm ÷ Osmolality
=12000÷290=41.3litres
Total body water of the given patient=12000÷265.7=45.1litres
Excess of water in the body of the given patient=45.1-41.3=3.8litres
B.Diuretics are a group of drugs which can be used for this patient whose mechanism of action is by inhibiting sodium reabsorption by the apical membrane Na,K,2Cl cotransporter in the ascending loop of Henle .
The Blood Pressure of the patient will fall when given diuretics and hence a fine balance has to be maintained between the dosage of the diuretic and the BP of the patient
The final osmolality of the patient would momentarily be brought to the normal range of 275mOsm to 290mOsm
This is a treatment to remove the excess of fluid from the body and that which reduces the venous return to the heart thereby reducing the work load on the heart nevertheless the underlying cause for the cardiac failure needs to be treated
C.The patient is experiencing a mild metabolic acidosis as pH is 7.31 the normal pH is 7.35-7.43
This gets corrected by a respiratory compensatory alkalosis by increase of PaO2
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