A blood sample is analyzed within a Sanz electrode. The voltage difference = 44.90 mV. What is the resultant pH of this blood?
a. 7.11 c. 7.57
b. 7.34 d. 8.18
ANS: C
A voltage of 61.5 mV is developed for every pH unit difference between the sample and the measuring electrode, which remains 6.84. Therefore, the voltage difference is 44.90 mV divided by 61.5 mV, or 0.73. The resultant pH can be calculated as 6.84 + 0.73 = 7.57.
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