Calculating pH of 0.01 M $Ca(OH)_2$ Solution

Calcium hydroxide ($Ca(OH)_2$) is a strong base. It dissociates completely in an aqueous solution.

Step 1: Determine Hydroxide Ion Concentration $[OH^-]$

The dissociation reaction is:

$Ca(OH)_2 \rightarrow Ca^{2+} + 2OH^-$

For every mole of $Ca(OH)_2$ that dissolves, two moles of hydroxide ions ($OH^-$) are produced.

Given the concentration of $Ca(OH)_2$ is 0.01 M, the concentration of $OH^-$ ions is:

$[OH^-] = 2 \times [Ca(OH)_2] = 2 \times 0.01 \, M = 0.02 \, M$

Step 2: Calculate pOH

The pOH is calculated using the formula:

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

Substituting the calculated $[OH^-]$:

$pOH = -\log_{10}(0.02)$

To simplify the logarithm:

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

Using logarithm properties ($\log(ab) = \log a + \log b$ and $\log(10^x) = x$):

$pOH = -(\log_{10} 2 + \log_{10} 10^{-2})$

$pOH = -(\log_{10} 2 - 2)$

Given $\log_{10} 2 = 0.3010$:

$pOH = -(0.3010 - 2) = -(-1.6990) = 1.6990$

Step 3: Calculate pH

At standard temperature (assumed to be 25°C), the relationship between pH and pOH is:

$pH + pOH = 14$

Solving for pH:

$pH = 14 - pOH$

$pH = 14 - 1.6990$

$pH = 12.3010$

Rounding to one decimal place, the pH is 12.3.