Problem 39
Question
The average \(\mathrm{pH}\) of normal arterial blood is \(7.40 .\) At normal body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{w}=2.4 \times 10^{-14} \cdot\) Calculate \(\left[\mathrm{H}^{+}\right]\) \(\left[\mathrm{OH}^{-}\right],\) and \(\mathrm{pOH}\) for blood at this temperature. $$ \begin{array}{llll} \hline \text { pH } & \text { pOH } & \text { [H }^{+} \text {] } & \text { [OH }^{-} \text {] } & \text { Acidic or basic? } \\ \hline 5.25 & & & \\ & 2.02 & & \\ & & 4.4 \times 10^{-10} M & & \\ & & & 8.5 \times 10^{-2} M & \\ \hline \end{array} $$
Step-by-Step Solution
Verified Answer
The concentrations of \(\mathrm{H}^{+}\), \(\mathrm{OH}^{-}\), and \(\mathrm{pOH}\) for blood at \(37^{\circ}\mathrm{C}\) are as follows:
\([\mathrm{H}^{+}] = 3.98 \times 10^{-8} \mathrm{M}\)
\([\mathrm{OH}^{-}] = 6.03\times 10^{-7} \mathrm{M}\)
\(\mathrm{pOH} = 6.22\)
The blood is slightly basic as the pH is slightly higher than 7.
1Step 1: Recall the relationship between pH, pOH, [H⁺], and [OH⁻]
The following equations describe the relationships between pH, pOH, [H⁺], and [OH⁻]:
1. pH = -log([H⁺])
2. pOH = -log([OH⁻])
3. pH + pOH = 14 (at 25°C) which is not the case for our problem
In this exercise, we have the value of \(K_w\) at the specific temperature of 37°C, so we'll have to use the equation:
4. \(K_w\) = [H⁺][OH⁻]
Now, let's start by calculating [H⁺] and [OH⁻] value.
2Step 2: Calculate [H⁺] value using the given pH
We know the pH of normal arterial blood is 7.40. To find [H⁺], we can use equation 1:
[H⁺] = \(10^{-\text{pH}}\)
[H⁺] = \(10^{-7.40}= \underline{3.98 \times 10^{-8} \mathrm{M}}\)
3Step 3: Calculate [OH⁻] value using the given Kw and [H⁺]
To find the [OH⁻] concentration, we can use equation 4:
\(K_w\) = [H⁺][OH⁻]
We know \(K_w = 2.4 \times 10^{-14}\) and [H⁺] \(= 3.98 \times 10^{-8} \mathrm{M}\). So,
\(2.4 \times 10^{-14} = (3.98 \times 10^{-8})([\mathrm{OH}^{-}])\)
Now, solve for [OH⁻]:
[OH⁻]= \(\frac{2.4 \times 10^{-14}}{3.98 \times 10^{-8}}= \underline{6.03\times 10^{-7} \mathrm{M}}\)
4Step 4: Calculate pOH for blood at this temperature
Now, we have the value of [OH⁻], so we can calculate the pOH value using equation 2:
pOH = -log([OH⁻])
pOH = -log(\(6.03\times 10^{-7}\)) = \(\underline{6.22}\)
5Step 5: Fill in the table
Finally, we can fill in the table with the values we calculated:
$$
\begin{array}{llll}
\hline \text { pH } & \text { pOH } & \text { [H }^{+} \text {] } & \text {
[OH }^{-} \text {] } & \text { Acidic or basic? } \\\
\hline 7.40 & 6.22 & 3.98 \times 10^{-8} \mathrm{M} & 6.03\times 10^{-7} \mathrm{M} & \text{Slightly basic} \\\
\hline
\end{array}
$$
Since the pH is slightly higher than 7, the blood is slightly basic.
Key Concepts
pH and pOH RelationshipsHydrogen Ion ConcentrationHydroxide Ion ConcentrationAcid-Base Balance in Blood
pH and pOH Relationships
Understanding the balance between pH and pOH is crucial when studying acid-base chemistry. The pH scale measures the acidity or alkalinity of a solution and ranges from 0 to 14. Values below 7 represent acidic conditions, while values above 7 indicate basic (or alkaline) conditions. At a neutral pH of 7, the solution is neither acidic nor basic. The pH and pOH of a solution are inversely related and their sum at 25°C is always 14. This relationship is given by the formula:
\[ \text{pH} + \text{pOH} = 14 \]However, this equation can change with temperature, as seen in blood pH calculation at body temperature (37°C), where we use the specific ion product of water (\(K_w\)) to find the relationship between pH and pOH.To find pOH from pH (and vice versa), we can use these simple calculations derived from the equations mentioned in the exercise:
\[ \text{pOH} = 14 - \text{pH} \]\[ \text{pH} = 14 - \text{pOH} \]These relationships are fundamental in solving many acid-base problems, including those involving blood chemistry.
\[ \text{pH} + \text{pOH} = 14 \]However, this equation can change with temperature, as seen in blood pH calculation at body temperature (37°C), where we use the specific ion product of water (\(K_w\)) to find the relationship between pH and pOH.To find pOH from pH (and vice versa), we can use these simple calculations derived from the equations mentioned in the exercise:
\[ \text{pOH} = 14 - \text{pH} \]\[ \text{pH} = 14 - \text{pOH} \]These relationships are fundamental in solving many acid-base problems, including those involving blood chemistry.
Hydrogen Ion Concentration
The hydrogen ion concentration (\([H^+]\)) in a solution is key to determining its pH. As the concentration of hydrogen ions increases, the solution becomes more acidic and the pH decreases. Conversely, a decrease in hydrogen ions makes the solution more basic and the pH rises. The pH of a solution is the negative logarithm (base 10) of the hydrogen ion concentration:
\[ \text{pH} = -\log([H^+]) \]This logarithmic scale means that a small change in hydrogen ion concentration can result in a large change in pH. For example, a blood pH of 7.40 corresponds to a hydrogen ion concentration of \(3.98 \times 10^{-8} \mathrm{M}\), which is crucial in the tight regulation of the body's acid-base balance. Small deviations in \([H^+]\) can lead to significant physiological impacts, underscoring the importance of maintaining a stable blood pH.
\[ \text{pH} = -\log([H^+]) \]This logarithmic scale means that a small change in hydrogen ion concentration can result in a large change in pH. For example, a blood pH of 7.40 corresponds to a hydrogen ion concentration of \(3.98 \times 10^{-8} \mathrm{M}\), which is crucial in the tight regulation of the body's acid-base balance. Small deviations in \([H^+]\) can lead to significant physiological impacts, underscoring the importance of maintaining a stable blood pH.
Hydroxide Ion Concentration
Just as hydrogen ions are critical for acidity, hydroxide ions (\([OH^-]\)) play a key role in the basicity of a solution. The hydroxide ion concentration can be calculated using the ion product of water (\(K_w\)), especially at temperatures different from standard room temperature (25°C). The calculation of \([OH^-]\) is a two-step process: first, you need the concentration of hydrogen ions, and then with the equation \( K_w = [H^+][OH^-] \) you can solve for \([OH^-]\). In the case of blood at 37°C, \(K_w\) was given as \(2.4 \times 10^{-14}\), which was used alongside \([H^+]\) to compute the hydroxide ion concentration. The calculated \([OH^-]\) gives insight into the basicity of the blood and helps maintain the necessary homeostasis for proper bodily functions.
Acid-Base Balance in Blood
The acid-base balance in blood is a fine-tuned physiological parameter that is essential for life. Blood typically has a slightly alkaline pH of about 7.35 to 7.45. Regulatory systems in the body constantly monitor and adjust the levels of acids and bases to maintain this narrow range. This balance is indicative of proper metabolic processes and is important for enzyme function, oxygen transport, and electrical activity in cells among other physiological processes.
Maintaining this balance involves a complex interplay between buffers, the respiratory system, and the kidneys. When the pH of blood shifts out of this range, it results in a state of either acidosis (too acidic) or alkalosis (too basic), both of which can be detrimental to health. Monitoring the concentration of hydrogen and hydroxide ions helps in understanding the blood's pH level and assessing the body's acid-base status, which is vital in the medical field for diagnosing and treating various conditions.
Maintaining this balance involves a complex interplay between buffers, the respiratory system, and the kidneys. When the pH of blood shifts out of this range, it results in a state of either acidosis (too acidic) or alkalosis (too basic), both of which can be detrimental to health. Monitoring the concentration of hydrogen and hydroxide ions helps in understanding the blood's pH level and assessing the body's acid-base status, which is vital in the medical field for diagnosing and treating various conditions.
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