This question tests knowledge about the normal physiological parameters of the human body, specifically the pH level of blood. Maintaining a stable blood pH is crucial for various bodily functions.
Correct Option: C) 7.4
The normal physiological pH of human blood is maintained very precisely within a narrow range of 7.35 to 7.45. A value of 7.4 is the average and most commonly cited pH for human blood.
To find the pH of a weak acid solution, we first need to determine the concentration of hydrogen ions (\([\text{H}^+]\)) produced by its dissociation. Since it's a weak acid, it only partially dissociates. We can use the acid dissociation constant (\(K_a\)) and the initial concentration of the acid to set up an equilibrium expression and solve for \([\text{H}^+]\). Finally, pH is calculated using the formula \(\text{pH} = -\log[\text{H}^+]\).
Write the dissociation equilibrium for the weak acid (HA):
Let HA be the weak acid. Its dissociation in water is:
\[ \text{HA} ightleftharpoons \text{H}^+ + \text{A}^- \]Set up an ICE (Initial, Change, Equilibrium) table:
Initial concentration of HA = \(0.1 \text{ M}\)
Initial concentrations of \(\text{H}^+\) and \(\text{A}^-\) = \(0\)
Let \(x\) be the concentration of HA that dissociates, which is also the concentration of \(\text{H}^+\) and \(\text{A}^-\) formed at equilibrium.
| HA | \(\text{H}^+\) | \(\text{A}^-\) | |
|---|---|---|---|
| Initial | \(0.1\) | \(0\) | \(0\) |
| Change | \(-x\) | \(+x\) | \(+x\) |
| Equilibrium | \(0.1 - x\) | \(x\) | \(x\) |
Write the expression for the acid dissociation constant (\(K_a\)):
\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]Substitute equilibrium concentrations into the \(K_a\) expression:
\[ K_a = \frac{(x)(x)}{(0.1 - x)} = \frac{x^2}{0.1 - x} \]Apply the approximation:
Given \(K_a = 10^{-5}\) and the initial concentration is \(0.1 \text{ M}\). Since \(K_a\) is very small compared to the initial concentration (\(K_a / [\text{HA}]_0 = 10^{-5} / 0.1 = 10^{-4} \ll 1\)), we can assume that \(x\) is much smaller than \(0.1\). Therefore, \(0.1 - x \approx 0.1\).
The equation becomes:
\[ 10^{-5} = \frac{x^2}{0.1} \]Solve for \(x\) (which is \([\text{H}^+]\)):
\[ x^2 = 10^{-5} \times 0.1 \]\[ x^2 = 10^{-6} \]\[ x = \sqrt{10^{-6}} \]\[ x = 10^{-3} \text{ M} \]Calculate the pH:
Since \(x = [\text{H}^+]\), we have \([\text{H}^+] = 10^{-3} \text{ M}\).
The pH is given by:
\[ \text{pH} = -\log[\text{H}^+] \]\[ \text{pH} = -\log(10^{-3}) \]\[ \text{pH} = 3 \]Correct Option: D) 3
This problem involves calculating the pH of a solution after dilution. We need to first determine the initial moles of HCl, then calculate the new concentration after dilution, and finally use the concentration of H⁺ ions to find the pH.
The initial volume of HCl is \(V_1 = 1 \text{ mL} = 0.001 \text{ L}\).
The initial concentration of HCl is \(M_1 = 0.1 \text{ M}\).
Number of moles of HCl (\(n\)) is given by:
\[ n = M_1 \times V_1 \] \[ n = 0.1 \text{ mol/L} \times 0.001 \text{ L} = 0.0001 \text{ mol} \]The solution is diluted to a final volume of \(V_2 = 100 \text{ mL} = 0.1 \text{ L}\).
Since the number of moles of HCl remains constant upon dilution, the new concentration (\(M_2\)) can be calculated as:
\[ M_2 = \frac{n}{V_2} \] \[ M_2 = \frac{0.0001 \text{ mol}}{0.1 \text{ L}} = 0.001 \text{ M} \]Alternatively, using the dilution formula \(M_1V_1 = M_2V_2\):
\[ 0.1 \text{ M} \times 1 \text{ mL} = M_2 \times 100 \text{ mL} \] \[ M_2 = \frac{0.1 \times 1}{100} = 0.001 \text{ M} \]HCl is a strong acid, meaning it dissociates completely in water:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]Therefore, the concentration of H⁺ ions is equal to the concentration of HCl:
\[ [\text{H}^+] = 0.001 \text{ M} = 1 \times 10^{-3} \text{ M} \]The pH is defined as the negative logarithm (base 10) of the H⁺ ion concentration:
\[ \text{pH} = -\log[\text{H}^+] \] \[ \text{pH} = -\log(1 \times 10^{-3}) \] \[ \text{pH} = -(-3) = 3 \]Correct Option: A) 3
The pH of a solution is a measure of its hydrogen ion concentration. The relationship between pH and hydrogen ion concentration \([\text{H}^+]\) is defined by a logarithmic scale. We will use the formula for pH to calculate the hydrogen ion concentration.
B) \(10^{-4} \text{ mol/L}\) is the correct hydrogen ion concentration for a solution with pH = 4.
The question asks for the relationship between pKa and Ka. This is a fundamental concept in acid-base chemistry, specifically related to the strength of an acid. The 'p' notation (as in pH, pOH, pKa) generally signifies the negative logarithm (base 10) of the quantity that follows.
Correct Option: D) pKa = -log Ka. This option correctly represents the definition of pKa as the negative base-10 logarithm of the acid dissociation constant (Ka).
The strength of an acid is inversely related to its pKa value. A lower pKa value indicates a stronger acid, while a higher pKa value indicates a weaker acid. This relationship is crucial for determining the strongest acid among the given options.
C) Acid with pKa = 1. This acid has the lowest pKa value among the options, indicating it is the strongest acid.