Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution

Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid.

The correct answer and explanation is :

To calculate the acid dissociation constant (Ka) for a monoprotic acid, we need to use the given pH value and concentration of the acid to find the concentrations of the dissociated species in the solution. Here’s a step-by-step breakdown of how to solve this:

Step 1: Understand the problem

  • The acid is monoprotic, meaning it dissociates in water according to the following equation:
    [
    HA \rightarrow H^+ + A^-
    ]
  • The initial concentration of the acid (HA) is 0.0192 M.
  • The pH of the solution is 2.53, which will allow us to find the concentration of (H^+) (hydrogen ions).

Step 2: Determine the concentration of (H^+) ions

The pH of the solution is given as 2.53. To find the concentration of (H^+), we use the equation:
[
pH = -\log[H^+]
]
Rearranging to solve for ([H^+]):
[
[H^+] = 10^{-pH} = 10^{-2.53}
]
Using a calculator:
[
[H^+] = 2.95 \times 10^{-3} \, M
]

Step 3: Set up an ICE table

We now know that the concentration of (H^+) at equilibrium is (2.95 \times 10^{-3}) M. We can use this information to set up an ICE (Initial, Change, Equilibrium) table for the dissociation of the acid:

SpeciesInitial (M)Change (M)Equilibrium (M)
HA0.0192-x0.0192 – x
H⁺0+xx
A⁻0+xx

From the ICE table, we know that (x = [H^+] = 2.95 \times 10^{-3}) M. So, the concentration of the dissociated acid (A⁻) is also (2.95 \times 10^{-3}) M.

Step 4: Calculate the Ka

The acid dissociation constant (K_a) is given by the expression:
[
K_a = \frac{[H^+][A^-]}{[HA]}
]
Substitute the values:
[
K_a = \frac{(2.95 \times 10^{-3})(2.95 \times 10^{-3})}{0.0192 – 2.95 \times 10^{-3}}
]
[
K_a = \frac{8.70 \times 10^{-6}}{0.01625}
]
[
K_a = 5.36 \times 10^{-4}
]

Final Answer:

The acid dissociation constant, (K_a), for the acid is approximately (5.36 \times 10^{-4}).

Explanation:

To solve for (K_a), we first used the given pH to calculate the concentration of (H^+) ions. From there, we constructed an ICE table to represent the dissociation process, determining the concentrations of the acid (HA), hydrogen ions (H⁺), and the conjugate base (A⁻) at equilibrium. Finally, we applied the formula for (K_a), which relates these concentrations, and calculated the acid dissociation constant.

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