**Chapter 18 Problem Set****Exercise 18.55 - Enhanced - with Feedback**---**Part A****Question:** What is the pH of blood plasma?**Instructions:** Express your answer to one decimal place.**Answer:**\[ \text{pH} = 7.4 \]**Explanation:**To find the pH of blood plasma, identify the acid and the base, and substitute their concentrations into the Henderson-Hasselbalch equation to calculate the pH. In this case, \( \text{H}_2\text{CO}_3 \) is the acid and \( \text{HCO}_3^- \) is the base.**Equation:**\[ \text{pH} = \text{p}K_{a1} + \log \left(\frac{\text{[base]}}{\text{[acid]}}\right) \]\[ = 6.1 + \log \left(\frac{0.024}{0.0012}\right) \]\[ = 7.4 \]The calculation is based on normal blood plasma concentrations of \(0.024\,M\) in \( \text{HCO}_3^- \) and \(0.0012\,M\) in \( \text{H}_2\text{CO}_3 \). (pKa1 for \( \text{H}_2\text{CO}_3 \) at body temperature is 6.1.)**Part B****Question:** If the volume of blood in a normal adult is 4.7 L, what mass of HCl could be neutralized by the buffering system in blood before the pH fell below 7.0 (which would result in death)?**Instructions:** Express your answer to two significant figures.**Answer:** \[ m_{\text{HCl}} = 0.28\,g \] ---### Buffering Capacity of Blood**Concept Overview:**Blood is buffered by carbonic acid and the bicarbonate ion. The normal concentrations are:- \(0.024 \, M\) in \( \text{HCO}_3^- \)- \(0.0012 \, M\) in \( \text{H}_2\text{CO}_3 \)The \(pK_a\) for \( \text{H}_2\text{CO}_3 \) at body temperature is 6.1.**Neutralization Reaction:**When \( \text{H}_3\text{O}^+ \) reacts with \( \text{HCO}_3^- \), it forms water and carbonic acid:\[\text{H}_3\text{O}^+ (\text{aq}) + \text{HCO}_3^- (\text{aq}) \rightarrow \text{H}_2\text{O} (l) + \text{H}_2\text{CO}_3 (\text{aq})\]**Table of Changes:**- **Before addition:** - \(\text{H}_3\text{O}^+ \approx 0.00 \text{ mol}\) - \(\text{HCO}_3^- = 0.11 \text{ mol}\)- **After addition:** - \(\text{H}_3\text{O}^+ = (0.11 - x) \text{ mol}\) - \(\text{H}_2\text{CO}_3 = (5.6 \times 10^{-3} + x) \text{ mol}\)The \( \text{pH} \) is 7.0 after the addition of HCl. Substitute into the Henderson-Hasselbalch equation:\[\text{pH} = pK_a + \log \left(\frac{[\text{base}]}{[\text{acid}]}\right)\]\[7.0 = 6.1 + \log \left(\frac{0.11 - x}{5.6 \times 10^{-3} + x}\right)\]**Solving for \(x\):**Rearranging the equation gives:\[x = 7.6 \times 10^{-

The objective of the question is to determine the mass of that could be neutralized before the pH rose above .Let's assume the moles of to be x, in this case, the acid reacts to give the base form.From part B, the Henderson equation is re-written as follows-The pH is given as .The value of The volume of blood is .Substituting the known values in the above equation as follows-The moles of comes out to be .The molar mass of is .The mass is calculated as follows-The mass of is .