1. As a dialysis patient, Michelle has a 4-h dialysis treatment three times a we

Question

1. As a dialysis patient, Michelle has a 4-h dialysis treatment three times a week. When she arrives at the dialysis clinic, her weight, temperature, and blood pressure are taken and blood tests are done to determine the level of electrolytes and urea in her blood. In the dialysis center, tubes to the dialyzer are connected to the catheter she has had implanted. Blood is then pumped out of her body, through the dialyzer where it is filtered, and returned to her body. As Michelle's blood flows through the dialyzer, electrolytes from the dialysate move into her blood, and waste products in her blood move into the dialysate, which is continually renewed. To achieve normal serum electrolyte levels, dialysate fluid contains sodium, chloride, and magnesium levels that are equal to serum concentrations. These electrolytes are removed from the blood only if their concentrations are higher than normal. Typically, in dialysis patients, the potassium ion level is higher than normal. Therefore, initial dialysis may start with a low concentration of potassium ion in the dialysate. During dialysis, excess fluid is removed by osmosis. A 4-h dialysis session requires at least 120 L of dialysis fluid. During dialysis, the electrolytes in the dialysate are adjusted until the electrolytes have the same levels as normal serum. Initially the dialysate solution prepared for Michelle's pre-dialysis blood tests shows that the electrolyte levels in her blood are as follows:

HCO3 24 mEq/L, K+ 6.0 mEq/L, Na+148 mEq/L, Ca2+ 3.0 mEq/L, Mg2+ 1.0 mEq/L, Cl 111.0 mEq/L.

A dialysis solution is prepared for Michelle that contains the following:

HCO3  25.0 mEq/L , K+  2.5 mEq/L , Na+  120.0 mEq/L , Ca2+  5.0 mEq/L , Mg2+  3.5 mEq/L , Cl  106.0, glucose 5.0%(m/v).

You may want to reference ( pages 352 - 360) Section 9.6 while completing this problem.

Part A

What is the total positive charge, in milliequivalents per liter, of the electrolytes in the dialysate fluid? Express the charge in milliequivalents per liter as an integer. mEq/L

Part B

What is the total negative charge, in milliequivalents per liter, of the electrolytes in the dialysate fluid?
Express the charge in milliequivalents per liter to one decimal place. mEq/L

Part C

What is the net total charge, in milliequivalents per liter, of the electrolytes in the dialysate fluid? Express your answer as an integer. mEq/L Part D

What is the osmolarity of the dialysate fluid? Express your answer using two decimal places. Osm

Which electrolyte concentrations are higher than normal serum values in Michelle's pre-dialysis blood test (see the table)?

Check all that apply.

Part B

Which electrolyte concentrations are lower than normal serum values in Michelle's pre-dialysis blood test (see the table)?

Check all that apply.

Part C

Which electrolytes need to be increased by dialysis for Michelle's blood serum?

Check all that apply.

Part D

Which electrolytes are decreased by dialysis for Michelle's blood serum?

Check all that apply.

3.

Part A

Drag each item to the appropriate bin.

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Net movement of water will be from side A to side B.

The solution volumes of side A and
side B will not change over time.

Side B has a higher
concentration of salt than side A.

Net movement of salt
will be from side B to side A.

The solution volume
of side A will decrease over time.

True

False

Na+ Ca2+ K+ Cl HCO3 Mg2+ Electrolytes in Blood Plasma and Selected Intravenous Solutions Normal Concentrations of lons (mEq/L) Normal Saline Lactated Ringer's Replacement Solution Blood Plasma 0.9% NaCl Solution Maintenance Solution (extracellular Maintains electrolytes Replaces fluid loss Hydration Purpose and fluids Replaces electrolytes Cations 154 135-145 130 Na 140 3.5-5.5 4.5-5.5 1.5-3.0 2+ Mg 154 137 158 Total Anions Acetate 47 40 154 109 95-105 103 HCO, 22-28 20 Lactate 28 1.8-2.3 15 HPO. Citrate 154 137 158 Total 75

Explanation / Answer

A) Dialysate is also called dialysis fluid so, as mentioned it contains K+  2.5 mEq/L , Na+  120.0 mEq/L , Ca2+  5.0 mEq/L, Mg2+  3.5 mEq/L

therefore, total positive charge = 2.5+120.0+5.0+3.5 = 131 mEq/L

B) HCO3  25.0 mEq/L , Cl  106.0

therefore, total negative charge = 25.0+106.0 = 131.0 mEq/L

C) net total charge = +131-131 = 0 mEq/L

D) osmolarity

HCO3  25.0 mEq/L , K+  2.5 mEq/L , Na+  120.0 mEq/L , Ca2+  5.0 mEq/L , Mg2+  3.5 mEq/L , Cl  106.0, glucose 5.0%(m/v).

step 1 converting mEq/l to mg/l by using the formula mEq/L X eq. wt = mg/L

HCO3  25.0 mEq/L = 1525mg/L

K+  2.5 mEq/L = 97.5mg/L

Na+  120.0 mEq/L = 2760mg/L

Ca2+  5.0 mEq/L = 100mg/L

Mg2+  3.5 mEq/L = 42mg/L

Cl  106.0 mEq/L = 3710mg/L

Step 2 converting mg/L into Mol/L

HCO3 = 0.025 Mol/L

K+ = 0.0025Mol/L

Na+ 0.12, Mol/L

Ca2+ 0.0025 Mol/L

Mg2+ 0.000175 Mol/L

Cl 0.106Mol/L

Step 3 multiply molarity by osmoles

HCO3 = 0.025 Mol/L X 2 = 0.05 osmoles/L

K+ = 0.0025Mol/L X 2 = 0.005osmoles/L

Na+ 0.12, Mol/L X2 = 0.24osmoles/L

Ca2+ 0.0025 Mol/L X3 = 0.0075osmoles/L

Mg2+ 0.000175 Mol/L X3 = 0.000525osmoles/L

Cl 0.106Mol/L X 2 =0.212osmoles/L

glucose = 5gm/100ml

=( 5 *1000)/ (180.2 *100) = 0.27 osmoles/L

Osmolarity = 0.05 +0.005 +0.24 +0.0075+ 0.000525+0.212+ 0.27 = 0.78 osmoles/L

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