4.3 Ionic Product of Water and Calculation of pH and pOH

This section explores the fundamental concept of the ionic product of water ($K_w$) and its vital role in calculating pH and pOH values, which are measures of acidity and basicity in aqueous solutions.

The Ionic Product of Water ($K_w$)

Pure water undergoes a slight self-ionization, producing hydronium ions (${\mathrm{H}}_3{\mathrm{O}}^{+}$) and hydroxide ions ($\mathrm{O}{\mathrm{H}}^{-}$). The equilibrium for this process is:

${\mathrm{H}}_2\mathrm{O}(\mathrm{l})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}(\mathrm{aq})+\mathrm{O}{\mathrm{H}}^{-}(\mathrm{aq})$

The ionic product of water, denoted as $K_w$, is the product of the molar concentrations of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ (often simplified as ${\mathrm{H}}^{+}$) and $\mathrm{O}{\mathrm{H}}^{-}$ ions in an aqueous solution.

$K_w=[{\mathrm{H}}_3{\mathrm{O}}^{+}][\mathrm{O}{\mathrm{H}}^{-}]$

• The unit of $K_w$ is $\mathrm{mo}{\mathrm{l}}^2\mathrm{d}{\mathrm{m}}^{-6}$.
• At room temperature ($298\mathrm{K}$ or $25^{\circ }\mathrm{C}$), in pure water, the concentrations of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ and $\mathrm{O}{\mathrm{H}}^{-}$ ions are equal: $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=[\mathrm{O}{\mathrm{H}}^{-}]=1.0\times 10^{-7}\mathrm{M}$.
• Therefore, the value of $K_w$ at $298\mathrm{K}$ is: $K_w=(1.0\times 10^{-7})(1.0\times 10^{-7})=1.0\times 10^{-14}$

Introduction to $\mathbf{p}{\mathbf{K}}_{\mathbf{w}}$

Similar to how pH is defined from $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$, $\mathrm{p}{\mathrm{K}}_{\mathrm{w}}$ is defined from $K_w$:

$\mathrm{p}{\mathrm{K}}_{\mathrm{w}}=-\log K_w$

Using the value of $K_w$ at $298\mathrm{K}$:

$\mathrm{p}{\mathrm{K}}_{\mathrm{w}}=-\log (1.0\times 10^{-14})=14$

Relationships involving $\mathbf{p}{\mathbf{K}}_{\mathbf{w}}$, $\mathbf{pH}$, $\mathbf{pOH}$, $\mathbf{p}{\mathbf{K}}_{\mathbf{a}}$, and $\mathbf{p}{\mathbf{K}}_{\mathbf{b}}$

The ionic product of water provides crucial relationships between acid-base properties.

Relationship between $\mathbf{p}{\mathbf{K}}_{\mathbf{a}}$ and $\mathbf{p}{\mathbf{K}}_{\mathbf{b}}$

The dissociation constant of an acid ($\mathrm{p}{\mathrm{K}}_{\mathrm{a}}$) and its conjugate base ($\mathrm{p}{\mathrm{K}}_{\mathrm{b}}$) are related through $\mathrm{p}{\mathrm{K}}_{\mathrm{w}}$.

$\mathrm{p}{\mathrm{K}}_{\mathrm{a}}+\mathrm{p}{\mathrm{K}}_{\mathrm{b}}=\mathrm{p}{\mathrm{K}}_{\mathrm{w}}=14(\text{at }298\mathrm{K})$

This relationship allows us to determine the strength of a conjugate base if the strength of its corresponding acid is known, and vice versa.

Relationship between $\mathbf{pH}$ and $\mathbf{pOH}$

The pH of a solution (a measure of its acidity) and pOH (a measure of its basicity) are also linked by $\mathrm{p}{\mathrm{K}}_{\mathrm{w}}$.

$\mathrm{pH}+\mathrm{pOH}=\mathrm{p}{\mathrm{K}}_{\mathrm{w}}=14(\text{at }298\mathrm{K})$

This equation is fundamental for calculating one value if the other is known, which is often necessary when dealing with strong bases (where $[\mathrm{O}{\mathrm{H}}^{-}]$ is directly available) or strong acids (where $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$ is directly available).

Worked Examples

Here are step-by-step solutions to common problems involving the ionic product of water and pH calculations.

Example 4.3: Calculating $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$ from $[\mathrm{O}{\mathrm{H}}^{-}]$

If the concentration of $\mathrm{NaOH}$ in a solution is $2.5\times 10^{-4}\mathrm{M}$, what is the concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ ion at $25^{\circ }\mathrm{C}$?

1. Given values:

   • $[\mathrm{O}{\mathrm{H}}^{-}]=2.5\times 10^{-4}\mathrm{M}$
   • Temperature = $25^{\circ }\mathrm{C}$ (which means $K_w=1.0\times 10^{-14}$)

2. Apply the formula: The ionic product of water relates $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$ and $[\mathrm{O}{\mathrm{H}}^{-}]$. $K_w=[{\mathrm{H}}_3{\mathrm{O}}^{+}][\mathrm{O}{\mathrm{H}}^{-}]$

3. Show calculation: Substitute the known values into the equation and solve for $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$. $1.0\times 10^{-14}=[{\mathrm{H}}_3{\mathrm{O}}^{+}][2.5\times 10^{-4}]$ $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=\frac{1.0\times 10^{-14}}{2.5\times 10^{-4}}$ $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=4.0\times 10^{-11}\mathrm{M}$

Example 4.4: Calculating $\mathbf{pH}$ from $[\mathrm{O}{\mathrm{H}}^{-}]$ (Method 1)

Calculate the pH value of a $0.001\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ solution of $\mathrm{NaOH}$ at $25^{\circ }\mathrm{C}$.

1. Given values:

   • $[\mathrm{O}{\mathrm{H}}^{-}]=0.001\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}=1.0\times 10^{-3}\mathrm{M}$
   • Temperature = $25^{\circ }\mathrm{C}$ (so $K_w=1.0\times 10^{-14}$)

2. Apply the formulas:

   • First, use $K_w=[{\mathrm{H}}_3{\mathrm{O}}^{+}][\mathrm{O}{\mathrm{H}}^{-}]$ to find $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$.
   • Then, use $\mathrm{pH}=-\log [{\mathrm{H}}_3{\mathrm{O}}^{+}]$ to find the pH.

3. Show calculation:

   • Calculate $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$: $1.0\times 10^{-14}=[{\mathrm{H}}_3{\mathrm{O}}^{+}][1.0\times 10^{-3}]$ $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=\frac{1.0\times 10^{-14}}{1.0\times 10^{-3}}$ $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=1.0\times 10^{-11}\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$
   • Calculate pH: $\mathrm{pH}=-\log [{\mathrm{H}}_3{\mathrm{O}}^{+}]$ $\mathrm{pH}=-\log [1.0\times 10^{-11}]$ $\mathrm{pH}=11$

Example 4.5: Calculating $\mathbf{pH}$ from $[\mathrm{O}{\mathrm{H}}^{-}]$ (Method 2)

Calculate the pH of a $5.0\times 10^{-5}\mathrm{M}$ solution of sodium hydroxide.

1. Given values:

   • $[\mathrm{O}{\mathrm{H}}^{-}]=5.0\times 10^{-5}\mathrm{M}$

2. Apply the formulas:

   • First, use $\mathrm{pOH}=-\log [\mathrm{O}{\mathrm{H}}^{-}]$ to find pOH.
   • Then, use $\mathrm{pH}+\mathrm{pOH}=14$ to find the pH.

3. Show calculation:

   • Calculate pOH: $\mathrm{pOH}=-\log [\mathrm{O}{\mathrm{H}}^{-}]$ $\mathrm{pOH}=-\log [5.0\times 10^{-5}]$ $\mathrm{pOH}=4.30$
   • Calculate pH: $\mathrm{pH}=14-\mathrm{pOH}$ $\mathrm{pH}=14-4.30$ $\mathrm{pH}=9.70$

Possible Questions/Answers

These questions test your understanding of pH, pOH, and the ionic product of water.

Q1: The concentration of hydroxide ion in a given solution of slaked lime ($\mathrm{Ca}(\mathrm{OH}{)}_2$) is $0.001\mathrm{M}$. Calculate the concentration of hydrogen ion in it.

A: For $\mathrm{Ca}(\mathrm{OH}{)}_2$, since it is a strong base, if the solution's hydroxide ion concentration is given as $0.001\mathrm{M}$, then:

• $[\mathrm{O}{\mathrm{H}}^{-}]=0.001\mathrm{M}=1.0\times 10^{-3}\mathrm{M}$
• Using $K_w=[{\mathrm{H}}_3{\mathrm{O}}^{+}][\mathrm{O}{\mathrm{H}}^{-}]$: $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=\frac{K_w}{[\mathrm{O}{\mathrm{H}}^{-}]}=\frac{1.0\times 10^{-14}}{1.0\times 10^{-3}}=1.0\times 10^{-11}\mathrm{M}$

Q2: An aqueous solution contains $2\times 10^{-3}\mathrm{M}$ of hydrogen ions (${\mathrm{H}}^{+}$). Calculate pOH of this solution.

A:

• Given $[{\mathrm{H}}^{+}]=2\times 10^{-3}\mathrm{M}$.
• First, calculate pH: $\mathrm{pH}=-\log [{\mathrm{H}}^{+}]=-\log (2\times 10^{-3})=2.70$
• Then, use the relationship $\mathrm{pH}+\mathrm{pOH}=14$: $\mathrm{pOH}=14-\mathrm{pH}=14-2.70=11.30$

(Alternative method: Calculate $[\mathrm{O}{\mathrm{H}}^{-}]$ first) $[\mathrm{O}{\mathrm{H}}^{-}]=\frac{K_w}{[{\mathrm{H}}^{+}]}=\frac{1.0\times 10^{-14}}{2\times 10^{-3}}=5.0\times 10^{-12}\mathrm{M}$ $\mathrm{pOH}=-\log [\mathrm{O}{\mathrm{H}}^{-}]=-\log (5.0\times 10^{-12})=11.30$