Call 911! It’s an Emergency!

1. Call 911! It’s an Emergency! The Restoration of a Disturbed Buffer System to Normal Blood pH Values The Pennsylvania State University Chem 294 By: Kristen Woznick • Cardiac Arrest • Normal metabolism • C6H12O6 + 6O2 6CO2 +6H20 • is interrupted when cardiac arrest occurs because the blood stops circulating. Oxygen becomes no longer present in the bloodstream; therefore, glucose is metabolized into lactic acid. • C6H12O6  2H3C-CH(OH)COOH • Lactic acid reacts with the weak base component in the blood to produce carbonic acid, which alters the weak acid: weak base ratio and therefore adjusts the patient’s pH in the blood. • C3H6O3 + HCO3- H2CO3 + C6H5O3 • What is a Buffer System? • A buffer system is composed of both a weak acid and a weak base paired in conjugation that withstands changes in pH. The acid contributes hydronium ions, H30+, which react with the base to produce both water and the conjugate acid. The base contributes OH-, hydroxide ions, which react with the conjugate acid to produce water and a conjugate base. This allows the pH of a solution to be determined by the hydrogen ion concentration. • pH = -log[H+] • pH Scale • We look at the Henderson-Hasselbalch equation • to determine the pH. [A-) represents the weak base concentration and [HA] is the weak acid concentration. Buffers undergo and uphold their conditions when they are stressed by dilution or the addition of a strong acid or strong base. Dilution does not affect the pH of a buffer solution because the ratio [A-]/ [HA] remains the same no matter the volume. Therefore, in this experiment, the buffers can be made by mixing the weak acid and weak base together and then diluting the solution to a certain volume, usually one liter. As a result, the buffer concentration can be determined. • Buffer concentration = [HA] + [A-] • Why is Blood a Buffer? • Blood is a buffer with a standard pH between 7.35 and 7.45 in order for the individual to carry oxygen. This buffer is comprised of H2CO3 and –HCO3 in a 1:20 ratio. If the ratio decreases, pH rises and individual becomes alkalotic. In this case, an acid component is added to the buffer to readjust the pH imbalance. If the ratio increases, pH decreases and individual becomes acidotic because there is excess carbon dioxide that cannot be excreted. As a result, the hydrogen ion concentration increases, decreasing the pH. • For further information, please reference: • 1.) Brown, Theodore L.; LeMay, H. Eugene; Bursten, Bruce E.; Murphy, Catherine J.; Woodward, Patrick. Chemistry – The Central Science. Pearson Education, Inc.: Upple Saddle River, NJ, 2009, pp. 675-735. • 2.) Wink, Donald J.; Gislason, Sharon Fetzer; Kuehn, Julie Ellefson. Working with Chemistry: A Laboratory Inquiry Program. W. H. Freeman and Company: New York, New York, 2010, pp. G1 – G22. What Do We Need to Do? The nurses need to determine the [A-] of the disturbed buffer of the unhealthy patient and compare this with the normal buffer of pH 7.35 of a normal and healthy patient. By comparing the concentration of the weak acid and weak base in a normal patient to a cardiac arrest patient, the nurse can determine the amount by which the concentration of [A-] needs to change. General Equation: [HA] + [A-] = X  [HA] = X – [A-] X= Molarity of Buffer Solution X = ? g NaH2PO4/ Liter x 1 Mole/ 119.99 g = ? M Then, using the Henderson-Hasselbalch equation 10(pKa-pH) = [HA]/ [A-]  10(pKa-pH) = (X/ [A-]) – 1 C1V1 = C2V2 C1 = standard concentration of hydrogen phosphate = 0.50 M V1 = amount of hydrogen phosphate that needs to be added C2 = {(normal ratio/ disturbed ratio) X [A- (disturbed]} – [A- (disturbed)] V2 = amount of buffer solution present Example Normal buffer solution (pH=7.35) X=3.00 Grams NaH2PO4/ Liter x 1 Mole/ 119.99 Grams = 0.0250 M 10(7.21-7.35) = (0.0250 M/ [A-]) – 1  [A-] = 0.0145 M [HA] = X – [A-]  [HA] = 0.025 M – 0.0145 M = 0.0105 M [A-]/ [HA] = 0.0145 M / 0.0105 M = 1.38 Disturbed buffer solution (pH=7.10) X = 0.0250 M 10(7.21-7.10) = (0.0250 M/ [A-]) – 1  [A-] = 0.0109 M [HA] = X – [A-]  [HA] = 0.0250 M – 0.0109 M = 0.0141 M [A-]/ [HA] = 0.0109M/ 0.0141M = 0.773 Normal ratio/ disturbed ratio = 1.38/ 0.773 = 1.785 1.785 x [A- (disturbed)]  1.785 x 0.0109 M = 0.0195 M 0.0195M – [A- (disturbed)]  0.0195M -0.0109M = 0.0086 M C1V1 = C2V2 C1 = 0.050 M C2 = 0.0086 M V1 = ? V2 = 50 mL (0.050M)(v1) = (0.0086M)(50mL)  V1 = 8.6 mL 8.6 mL of hydrogen phosphate are added to 50 mL of the phosphate buffer solution with a pH of 7.10 resulting in a solution with a pH in the normal range of 7.35 to 7.45. Results Conclusion In general, the amount of “bicarb” that needs to be added to a patient that undergoes cardiac arrest depends upon many factors including blood pH, volume, etc… By using the general formula, it can be determined how much “bicarb” needs to be added to solutions that have a wide range of pH values and volumes as seen in the chart above. In the end, the pH imbalance is corrected and the patient has been saved! Additional work could be done to create a reference table that determines the amount of “bicarb” to be added to patients presenting varying factors. Hypothetical Situation In a clinical scenario, a patient enters a hospital emergency room after experiencing cardiac arrest. Even though the medics gave the victim CPR, further treatment was still needed to prevent the patient from experiencing shock and potential death. The nurses need to quickly administer “bicarb” to the patient’s blood which has a reduced pH between 7.00 and 7.25. This will allow the blood to regain a value in the normal pH range, 7.35-7.45, but they do not know how much to add to the patient’s blood. Therefore, the goal of this experiment is to determine how to quickly calculate the necessary amount of “bicarb” that is to be added to a cardiac arrest patient’s blood if the volume, disturbed pH value, and other important factors are known so the patient can be saved. What Do We Need? It is known that the carbonic acid – hydrogen carbonate buffer easily emits carbon dioxide into the atmosphere; therefore, it is not suitable for laboratory work. Instead, the H2PO4- - HPO42- buffer system is used in this experiment due to the similar properties. One liter 0.025 M buffer solutions were made by dissolving sodium dihydrogen phosphate in distilled water . The pH values needed to be adjusted to 7.30, 7.25, 7.20, 7.15, and 7.10 by adding the appropriate amount of 0.50 M NaOH. The Buffers were readjusted to the normal and healthy pH range of 7.35-7.45 by adding an unique amount of 0.050 M HPO42-. From now on, only the buffer with disturbed pH of 7.10 is explained in thorough detail; however, the same method and general equations are utilized for the variety of pH values within the range of 7.10 to 7.25 and volumes of 50 and 100 mL. How Do We Make the Buffers? Acknowledgment: Thank you to Jing Dong , Mary Shoemaker, the stockroom staff, and especially Dr. Keiser for allowing me to obtain this valuable sophomore internship.