The pH of `10^(-10)` M NaOH solution lies between........and ........... .

Answer

To determine the pH of a `10^(-10)` M NaOH solution, we can follow these steps:

### Step 1: Understand the nature of NaOH
NaOH is a strong base, which means it completely dissociates in water to produce hydroxide ions (OH⁻).

### Step 2: Calculate the concentration of OH⁻ from NaOH
Given that the concentration of NaOH is `10^(-10)` M, the concentration of OH⁻ from NaOH will also be `10^(-10)` M.

### Step 3: Consider the contribution of OH⁻ from water
In pure water, the concentration of OH⁻ is `10^(-7)` M due to the self-ionization of water. Since `10^(-10)` M NaOH is a very dilute solution, we need to consider the contribution of OH⁻ from both NaOH and water.

### Step 4: Calculate the total concentration of OH⁻
The total concentration of OH⁻ will be the sum of the OH⁻ from NaOH and the OH⁻ from water:
\[ [\text{OH}^-]_{\text{total}} = [\text{OH}^-]_{\text{NaOH}} + [\text{OH}^-]_{\text{water}} = 10^{-10} + 10^{-7} \]
Since `10^(-7)` M is much larger than `10^(-10)` M, we can approximate:
\[ [\text{OH}^-]_{\text{total}} \approx 10^{-7} \text{ M} \]

### Step 5: Calculate the total concentration of OH⁻ more accurately
To be more precise, we can factor out `10^(-7)`:
\[ [\text{OH}^-]_{\text{total}} = 10^{-7} (1 + 10^{-3}) \approx 10^{-7} (1.001) \approx 1.001 \times 10^{-7} \text{ M} \]

### Step 6: Calculate the pOH
Using the formula for pOH:
\[ \text{pOH} = -\log([\text{OH}^-]) = -\log(1.001 \times 10^{-7}) \]
This can be approximated as:
\[ \text{pOH} \approx 7 - \log(1.001) \approx 7 - 0.0004 \approx 6.9996 \]

### Step 7: Calculate the pH
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
Thus,
\[ \text{pH} = 14 - \text{pOH} \approx 14 - 6.9996 \approx 7.0004 \]

### Step 8: Determine the range of pH
Since the calculated pH is approximately 7.0004, we can conclude that the pH of the `10^(-10)` M NaOH solution lies between 7 and 8.

### Final Answer
The pH of `10^(-10)` M NaOH solution lies between **7 and 8**.

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