An aquoeous solution twice alkaline as water, The pH of the solution is near to

Options

A. (a)6.7
B. (b)6.4
C. (c)7.3
D. (d)7.6

Answer

To determine the pH of an aqueous solution that is twice as alkaline as water, we can follow these steps:

### Step 1: Understand the concept of alkalinity
Alkalinity refers to the capacity of a solution to neutralize acids, which is often associated with the concentration of hydroxide ions (OH⁻). Pure water has a pH of 7, which means it has equal concentrations of hydrogen ions (H⁺) and hydroxide ions (OH⁻).

### Step 2: Determine the concentration of hydroxide ions in water
At 25°C, the ion product of water (Kw) is given by:
\[ K_w = [H^+][OH^-] = 10^{-14} \]
In neutral water, the concentration of H⁺ is equal to the concentration of OH⁻:
\[ [H^+] = [OH^-] = 10^{-7} \, \text{M} \]

### Step 3: Calculate the concentration of hydroxide ions in the solution
The problem states that the solution is "twice alkaline" compared to water. This means the concentration of hydroxide ions in the solution is:
\[ [OH^-] = 2 \times 10^{-7} \, \text{M} \]

### Step 4: Calculate the pOH of the solution
To find the pOH, we use the formula:
\[ pOH = -\log[OH^-] \]
Substituting the concentration:
\[ pOH = -\log(2 \times 10^{-7}) \]
Using properties of logarithms:
\[ pOH = -\log(2) - \log(10^{-7}) \]
\[ pOH = -0.301 - (-7) \]
\[ pOH = 7 - 0.301 = 6.699 \]

### Step 5: Calculate the pH of the solution
We know that:
\[ pH + pOH = 14 \]
Thus, we can find the pH:
\[ pH = 14 - pOH \]
Substituting the pOH value we calculated:
\[ pH = 14 - 6.699 \]
\[ pH = 7.301 \]

### Conclusion
The pH of the solution that is twice as alkaline as water is approximately 7.3.

---