The `P^(H)` of a dibasic acid is 3.699. Its molarity is

Options

A. `2xx10^(-4)` M
B. `4xx10^(-4)` M
C. `2xx10^(-3)`M
D. `1xx10^(-4)` M

Answer

To find the molarity of a dibasic acid given its pH, we can follow these steps:

### Step 1: Understand the relationship between pH and [H⁺]
The pH of a solution is defined as:
\[ \text{pH} = -\log[H^+] \]
Where \([H^+]\) is the concentration of hydrogen ions in moles per liter (M).

### Step 2: Calculate [H⁺] from pH
Given that the pH of the dibasic acid is 3.699, we can calculate the concentration of hydrogen ions:
\[ [H^+] = 10^{-\text{pH}} \]
Substituting the given pH value:
\[ [H^+] = 10^{-3.699} \]

### Step 3: Calculate the numerical value of [H⁺]
Using a calculator:
\[ [H^+] \approx 2.00 \times 10^{-4} \, \text{M} \]

### Step 4: Relate [H⁺] to the molarity of the dibasic acid
Since the acid is dibasic, it can donate two protons (H⁺) for every molecule of acid. Therefore, the concentration of the acid \([HA]\) is half the concentration of hydrogen ions:
\[ [HA] = \frac{[H^+]}{2} \]

### Step 5: Calculate the molarity of the dibasic acid
Substituting the value of \([H^+]\):
\[ [HA] = \frac{2.00 \times 10^{-4}}{2} = 1.00 \times 10^{-4} \, \text{M} \]

### Final Answer
The molarity of the dibasic acid is:
\[ \text{Molarity} = 1.00 \times 10^{-4} \, \text{M} \]

---