the pH if .60M amomonium chloride,NH4Cl is 4.74. what isthe valuke of Ka for this acid

To find the value of \( K_a \) for ammonium chloride (NH4Cl), we first need to understand the relationship between the pH of a solution and the dissociation constant of the weak acid it produces. Ammonium chloride is a salt that dissociates in water to form ammonium ions (NH4+) and chloride ions (Cl-). The ammonium ion can act as a weak acid in solution, which is what we are interested in here.

Understanding the Relationship Between pH and \( K_a \)

The pH of a solution is a measure of its acidity. In this case, the pH is given as 4.74. To find the \( K_a \) value, we can follow these steps:

Step 1: Calculate the Concentration of Hydrogen Ions

First, we need to convert the pH to the concentration of hydrogen ions \([H^+]\) using the formula:

\([H^+] = 10^{-\text{pH}}\

Substituting the given pH:

\([H^+] = 10^{-4.74} \approx 1.82 \times 10^{-5} \, \text{M}\

Step 2: Set Up the Equilibrium Expression

When NH4+ dissociates in water, it can be represented by the following equilibrium reaction:

NH4+ ⇌ H+ + NH3

The equilibrium constant expression for this reaction is:

\( K_a = \frac{[H^+][NH3]}{[NH4+]}\)

Step 3: Determine the Concentrations at Equilibrium

Initially, the concentration of NH4+ is 0.60 M. As the reaction proceeds, let \( x \) be the amount that dissociates. At equilibrium, we have:

• \([NH4^+] = 0.60 - x\)
• \([H^+] = x\)
• \([NH3] = x\)

Since we calculated \([H^+]\) to be approximately \( 1.82 \times 10^{-5} \, \text{M}\), we can set \( x = 1.82 \times 10^{-5} \, \text{M}\).

Step 4: Substitute Values into the \( K_a \) Expression

Now we can substitute these values into the \( K_a \) expression:

\( K_a = \frac{(1.82 \times 10^{-5})(1.82 \times 10^{-5})}{0.60 - 1.82 \times 10^{-5}} \)

Since \( 1.82 \times 10^{-5} \) is very small compared to 0.60, we can approximate:

\( K_a \approx \frac{(1.82 \times 10^{-5})^2}{0.60} \)

Step 5: Calculate \( K_a \)

Now, calculating \( K_a \):

\( K_a \approx \frac{3.31 \times 10^{-10}}{0.60} \approx 5.52 \times 10^{-10} \)

Final Result

Thus, the value of \( K_a \) for the ammonium ion in ammonium chloride solution is approximately \( 5.52 \times 10^{-10} \). This indicates that NH4+ is a weak acid, as expected, since it has a low dissociation constant.