NaOH is a strong base. What will be pH of $$5.0 \times 10^{-2}$$ M NaOH solution? (log 2 = 0.3)

NaOH is a strong base, so it dissociates completely in water. The dissociation reaction is:

$$\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-$$

Given the concentration of NaOH is $$5.0 \times 10^{-2}$$ M, the concentration of hydroxide ions, $$[\text{OH}^-]$$, is also $$5.0 \times 10^{-2}$$ M.

To find the pH, we first need to find the pOH. The pOH is defined as the negative logarithm (base 10) of the hydroxide ion concentration:

$$\text{pOH} = -\log_{10} [\text{OH}^-]$$

Substituting the value:

$$\text{pOH} = -\log_{10} (5.0 \times 10^{-2})$$

Using the logarithm property $$\log(a \times b) = \log a + \log b$$, we can split this:

$$\text{pOH} = -\left[ \log_{10} 5.0 + \log_{10} (10^{-2}) \right]$$

We know that $$\log_{10} (10^{-2}) = -2$$. Now, $$\log_{10} 5.0$$ can be expressed as $$\log_{10} \left(\frac{10}{2}\right) = \log_{10} 10 - \log_{10} 2$$. Since $$\log_{10} 10 = 1$$ and given that $$\log_{10} 2 = 0.3$$, we have:

$$\log_{10} 5.0 = 1 - 0.3 = 0.7$$

Substituting back:

$$\text{pOH} = -\left[ 0.7 + (-2) \right] = -\left[ 0.7 - 2 \right] = -\left[-1.3\right]$$

Simplifying:

$$\text{pOH} = -(-1.3) = 1.3$$

Now, recall that pH and pOH are related by:

$$\text{pH} + \text{pOH} = 14$$

Solving for pH:

$$\text{pH} = 14 - \text{pOH} = 14 - 1.3 = 12.7$$

Hence, the pH of the solution is 12.7.

Comparing with the options:

A. 14.00

B. 13.70

C. 13.00

D. 12.70

The correct answer is Option D.