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Chapter 16: Problem 37

The average \(\mathrm{pH}\) of normal arterial blood is 7.40 . At normal body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{w}=2.4 \times 10^{-14} .\) Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\), and \(\mathrm{pOH}\) for blood at this temperature.

Short Answer

Expert verified
In summary, at a body temperature of 37°C and an average arterial blood pH of 7.40, the concentration of H+ ions in blood is \(1.00 \times 10^{-7.40}\) mol/L, the concentration of OH- ions is \(2.41 \times 10^{-6.60}\) mol/L, and the pOH value is approximately 6.60.

Step by step solution

01

Finding the concentration of H+ ions [H+] using pH

We are given the pH of blood, which is 7.40. We can use the equation pH = -log[H+] to find the concentration of H+ ions: 7.40 = -log[H+] To get [H+], we need to take the antilog of both sides of the equation: [H+] = 10^(-7.40) Calculate the value: [H+] = \(1.00 \times 10^{-7.40}\) mol/L So, the concentration of H+ ions in the blood is \(1.00 \times 10^{-7.40}\) mol/L.
02

Finding the concentration of OH- ions [OH-] using Kw

Now that we have the concentration of H+ ions, we can find the concentration of OH- ions using the ion product constant of water, Kw: Kw = [H+] * [OH-] We are given Kw = \(2.4 \times 10^{-14}\), and [H+] = \(1.00 \times 10^{-7.40}\) mol/L. Plugging these values into the equation: \(2.4 \times 10^{-14}\) = \(1.00 \times 10^{-7.40}\) * [OH-] Solve for [OH-]: [OH-] = \(\frac{2.4 \times 10^{-14}}{1.00 \times 10^{-7.40}}\) mol/L Calculate the value: [OH-] = \(2.41 \times 10^{-6.60}\) mol/L So, the concentration of OH- ions in the blood is \(2.41 \times 10^{-6.60}\) mol/L.
03

Finding the pOH value

Now that we have found the concentration of OH- ions, we can calculate the pOH value for blood using the equation pOH = -log[OH-]: pOH = -log[\(2.41 \times 10^{-6.60}\)] Calculate the value: pOH ≈ 6.60 So, the pOH of blood at this temperature is approximately 6.60. In summary, at a body temperature of 37°C and an average arterial blood pH of 7.40, the concentration of H+ ions in blood is \(1.00 \times 10^{-7.40}\) mol/L, the concentration of OH- ions is \(2.41 \times 10^{-6.60}\) mol/L, and the pOH value is approximately 6.60.

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Most popular questions from this chapter

A particular sample of vinegar has a pH of 2.20 . If acetic acid is the only acid that vinegar contains \(\left(K_{a}=1.8 \times 10^{-5}\right)\), calculate the concentration of acetic acid in the vinegar.

Write the chemical equation and the \(K_{a}\) expression for the dissociation of each of the following acids in aqueous solution. First show the reaction with \(\mathrm{H}^{+}(a q)\) as a product and then with the hydronium ion: \((\mathbf{a}) \mathrm{HNO}_{2},\) (b) \(\mathrm{ClH}_{2} \mathrm{CCOOH}\).

\(\mathrm{NH}_{3}(g)\) and \(\mathrm{HCl}(g)\) react to form the ionic solid \(\mathrm{NH}_{4} \mathrm{Cl}(s) .\) Which substance is the Brønsted-Lowry acid in this reaction? Which is the Brønsted-Lowry base?

A solution is made by adding $1.000 \mathrm{~g} \mathrm{Ca}(\mathrm{OH})_{2}(s), 100.0 \mathrm{~mL}$ of \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\), and enough water to make a final volume of \(350.0 \mathrm{~mL}\). Assuming that all of the solid dissolves, what is the \(\mathrm{pH}\) of the final solution?

Succinic acid $\left(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{4}\right),\( which we will denote \)\mathrm{H}_{2} \mathrm{Suc}$ is a biologically relevant diprotic acid with the structure shown below. At \(25^{\circ} \mathrm{C}\), the acid-dissociation constants for succinic acid are \(K_{a 1}=6.9 \times 10^{-5}\) and \(K_{a 2}=2.5 \times 10^{-6} .\) (a) Determine the pH of a \(0.32 \mathrm{M}\) solution of $\mathrm{H}_{2} \mathrm{Suc}\( at \)25^{\circ} \mathrm{C}$, assuming that only the first dissociation is relevant. (b) Determine the molar concentration of \(\mathrm{Suc}^{2-}\) in the solution in part (a). (c) Is the assumption you made in part (a) justified by the result from part (b)? (d) Will a solution of the salt NaHSuc be acidic, neutral, or basic?
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