# 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 ($\mathrm{H_3O^+}$) and hydroxide ions ($\mathrm{OH^-}$). The equilibrium for this process is: $$ \mathrm{H_2O (l) + H_2O (l) \rightleftharpoons H_3O^+ (aq) + OH^- (aq)} $$ The ionic product of water, denoted as $K_w$, is the product of the molar concentrations of $\mathrm{H_3O^+}$ (often simplified as $\mathrm{H^+}$) and $\mathrm{OH^-}$ ions in an aqueous solution. $$ K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] $$ - The unit of $K_w$ is $\mathrm{mol^2\,dm^{-6}}$. - At room temperature ($298\ \mathrm{K}$ or $25^\circ\mathrm{C}$), in pure water, the concentrations of $\mathrm{H_3O^+}$ and $\mathrm{OH^-}$ ions are equal: $[\mathrm{H_3O^+}] = [\mathrm{OH^-}] = 1.0 \times 10^{-7}\ \mathrm{M}$. - 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 $\mathbf{pK_w}$ Similar to how pH is defined from $[\mathrm{H_3O^+}]$, $\mathrm{pK_w}$ is defined from $K_w$: $$ \mathrm{pK_w} = -\log K_w $$ Using the value of $K_w$ at $298\ \mathrm{K}$: $$ \mathrm{pK_w} = -\log (1.0 \times 10^{-14}) = 14 $$ ## Relationships involving $\mathbf{pK_w}$, $\mathbf{pH}$, $\mathbf{pOH}$, $\mathbf{pK_a}$, and $\mathbf{pK_b}$ The ionic product of water provides crucial relationships between acid-base properties. ### Relationship between $\mathbf{pK_a}$ and $\mathbf{pK_b}$ The dissociation constant of an acid ($\mathrm{pK_a}$) and its conjugate base ($\mathrm{pK_b}$) are related through $\mathrm{pK_w}$. $$ \mathrm{pK_a} + \mathrm{pK_b} = \mathrm{pK_w} = 14 \quad (\text{at } 298\ \mathrm{K}) $$ 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 $\mathrm{pK_w}$. $$ \mathrm{pH} + \mathrm{pOH} = \mathrm{pK_w} = 14 \quad (\text{at } 298\ \mathrm{K}) $$ This equation is fundamental for calculating one value if the other is known, which is often necessary when dealing with strong bases (where $[\mathrm{OH^-}]$ is directly available) or strong acids (where $[\mathrm{H_3O^+}]$ 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 $[\mathrm{H_3O^+}]$ from $[\mathrm{OH^-}]$ If the concentration of $\mathrm{NaOH}$ in a solution is $2.5 \times 10^{-4}\ \mathrm{M}$, what is the concentration of $\mathrm{H_3O^+}$ ion at $25^\circ\mathrm{C}$? 1. **Given values:** - $[\mathrm{OH^-}] = 2.5 \times 10^{-4}\ \mathrm{M}$ - Temperature = $25^\circ\mathrm{C}$ (which means $K_w = 1.0 \times 10^{-14}$) 2. **Apply the formula:** The ionic product of water relates $[\mathrm{H_3O^+}]$ and $[\mathrm{OH^-}]$. $$ K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] $$ 3. **Show calculation:** Substitute the known values into the equation and solve for $[\mathrm{H_3O^+}]$. $$ 1.0 \times 10^{-14} = [\mathrm{H_3O^+}][2.5 \times 10^{-4}] $$ $$ [\mathrm{H_3O^+}] = \frac{1.0 \times 10^{-14}}{2.5 \times 10^{-4}} $$ $$ [\mathrm{H_3O^+}] = 4.0 \times 10^{-11}\ \mathrm{M} $$ ### Example 4.4: Calculating $\mathbf{pH}$ from $[\mathrm{OH^-}]$ (Method 1) Calculate the pH value of a $0.001\ \mathrm{mol\,dm^{-3}}$ solution of $\mathrm{NaOH}$ at $25^\circ\mathrm{C}$. 1. **Given values:** - $[\mathrm{OH^-}] = 0.001\ \mathrm{mol\,dm^{-3}} = 1.0 \times 10^{-3}\ \mathrm{M}$ - Temperature = $25^\circ\mathrm{C}$ (so $K_w = 1.0 \times 10^{-14}$) 2. **Apply the formulas:** - First, use $K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}]$ to find $[\mathrm{H_3O^+}]$. - Then, use $\mathrm{pH} = -\log[\mathrm{H_3O^+}]$ to find the pH. 3. **Show calculation:** - Calculate $[\mathrm{H_3O^+}]$: $$ 1.0 \times 10^{-14} = [\mathrm{H_3O^+}][1.0 \times 10^{-3}] $$ $$ [\mathrm{H_3O^+}] = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-3}} $$ $$ [\mathrm{H_3O^+}] = 1.0 \times 10^{-11}\ \mathrm{mol\,dm^{-3}} $$ - Calculate pH: $$ \mathrm{pH} = -\log[\mathrm{H_3O^+}] $$ $$ \mathrm{pH} = -\log[1.0 \times 10^{-11}] $$ $$ \mathrm{pH} = 11 $$ ### Example 4.5: Calculating $\mathbf{pH}$ from $[\mathrm{OH^-}]$ (Method 2) Calculate the pH of a $5.0 \times 10^{-5}\ \mathrm{M}$ solution of sodium hydroxide. 1. **Given values:** - $[\mathrm{OH^-}] = 5.0 \times 10^{-5}\ \mathrm{M}$ 2. **Apply the formulas:** - First, use $\mathrm{pOH} = -\log[\mathrm{OH^-}]$ to find pOH. - Then, use $\mathrm{pH} + \mathrm{pOH} = 14$ to find the pH. 3. **Show calculation:** - Calculate pOH: $$ \mathrm{pOH} = -\log[\mathrm{OH^-}] $$ $$ \mathrm{pOH} = -\log[5.0 \times 10^{-5}] $$ $$ \mathrm{pOH} = 4.30 $$ - 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 ($\mathrm{Ca(OH)_2}$) is $0.001\ \mathrm{M}$. Calculate the concentration of hydrogen ion in it. **A:** For $\mathrm{Ca(OH)_2}$, since it is a strong base, if the solution's hydroxide ion concentration is given as $0.001\ \mathrm{M}$, then: - $[\mathrm{OH^-}] = 0.001\ \mathrm{M} = 1.0 \times 10^{-3}\ \mathrm{M}$ - Using $K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}]$: $$ [\mathrm{H_3O^+}] = \frac{K_w}{[\mathrm{OH^-}]} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-3}} = 1.0 \times 10^{-11}\ \mathrm{M} $$ **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 $[\mathrm{OH^-}]$ first) $$ [\mathrm{OH^-}] = \frac{K_w}{[\mathrm{H^+}]} = \frac{1.0 \times 10^{-14}}{2 \times 10^{-3}} = 5.0 \times 10^{-12}\ \mathrm{M} $$ $$ \mathrm{pOH} = -\log[\mathrm{OH^-}] = -\log(5.0 \times 10^{-12}) = 11.30 $$

4.3 Ionic Product Of Water And Calculation Of Mathbf P H And Mathbf P O H | Acid Base Chemistry | Chemistry 12