Q.The pH of 0.1M monobasic acid is 4.50. Calculate the concentration of species H+, A– and HA at equilibrium. Also, determine the value of Ka and pKa of the monobasic acid.

The key idea is that pH directly gives the hydrogen ion concentration for a monobasic acid, where each molecule dissociates to give one $\text{H}^+$ and one $\text{A}^-$; once $[\text{H}^+]$ is known, $K_a$ follows from the equilibrium expression.

Step 1: Use the definition of pH:

Step 2: Substitute the given pH and solve for $[\text{H}^+]$:

Step 3: Compute the value:

Step 4: For a monobasic acid, $[\text{H}^+] = [\text{A}^-]$, so both species have the same concentration.

Step 5: Find $[\text{HA}]$ at equilibrium. The initial concentration was 0.1 M and only $3.16\times10^{-5}$ M dissociated -- negligible against 0.1 M: …

The pH scale is a logarithmic measure of hydrogen ion concentration. The relationship is:

So if you know the pH, you can always find $[\text{H}^+]$ by taking the antilog. For a monobasic acid HA, the dissociation is simple:

Every molecule of HA that dissociates gives one $\text{H}^+$ and one $\text{A}^-$. Therefore, in the solution, the concentration of $\text{H}^+$ equals the concentration of $\text{A}^-$, provided there is no other significant source of $\text{H}^+$.

Let's work through it step by step.

1. Find $[\text{H}^+]$ from the given pH. The pH is 4.50. Using the definition:

To evaluate $10^{-4.50}$, note that $10^{-4.50} = 10^{-5} \times 10^{0.50}$. Since $10^{0.50} = \sqrt{10} \approx 3.1623$, we get:

2. Relate $[\text{H}^+]$ to $[\text{A}^-]$. From the dissociation equation, each $\text{H}^+$ comes from one HA molecule, producing one $\text{A}^-$. So at equilibrium:

3. Find $[\text{HA}]$ at equilibrium. The acid started at 0.1 M, and only $3.16\times10^{-5}$ M of it dissociated -- about 0.03% -- so this loss is negligible against the initial concentration:

4. Compute the ionization constant $K_a$. Substituting the equilibrium concentrations into the equilibrium expression for $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$: …