Because the concentration of hydrogen ions in aqueous solution varies by many orders of magnitude (from over 1.0 mol dm^-3 to less than 10^-14 mol dm^-3), working with raw concentrations can be awkward. To make these numbers easier to manage, we use the logarithmic pH scale, which converts concentration values into a range typically between 0 and 14.
The negative logarithm to the base 10 of the hydrogen ion concentration in aqueous solution:
\( \text{pH} = -\log_{10}[\text{H}^+] \)
🔑 Key Principle
To convert from pH back to hydrogen ion concentration, we rearrange the definition to:
\( [\text{H}^+] = 10^{-\text{pH}} \)
Because of the base-10 logarithmic scale, a change of exactly 1 pH unit represents a 10-fold change in hydrogen ion concentration.
pH of Strong Acids
Because strong acids dissociate completely in water, the concentration of hydrogen ions in a monoprotic strong acid (like \( \text{HCl} \) or \( \text{HNO}_3 \)) is exactly equal to the concentration of the acid itself:
\( [\text{H}^+] = [\text{Acid}] \)
For a diprotic strong acid like sulfuric acid (\( \text{H}_2\text{SO}_4 \)), each mole of acid releases two moles of protons:
\( [\text{H}^+] = 2 \times [\text{Acid}] \)
Once \( [\text{H}^+] \) is determined, you calculate pH directly using the pH formula.
The Ionic Product of Water (\( \text{K}_w \))
Water dissociates very slightly to establish a reversible equilibrium:
\( \text{H}_2\text{O(l)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \)
The equilibrium constant expression for this dissociation would include the concentration of water in the denominator. However, because water is in vast excess, its concentration remains virtually constant. We combine this constant value with the equilibrium constant to define a new constant, the ionic product of water:
The equilibrium constant for the self-ionisation of water, defined as:
\( \text{K}_w = [\text{H}^+\text{(aq)}][\text{OH}^-\text{(aq)}] \)
At 298 K, its value is exactly \( 1.00 \times 10^{-14}\text{ mol}^2\text{ dm}^{-6} \).
Because pure water contains equal concentrations of hydrogen and hydroxide ions, it is neutral. Thus, in pure water:
\( [\text{H}^+] = [\text{OH}^-] = \sqrt{\text{K}_w} \)
The dissociation of water is endothermic. According to Le Chatelier's principle, if temperature is increased, the equilibrium shifts to the right to absorb heat. This increases the concentration of both \( \text{H}^+ \) and \( \text{OH}^- \) ions, raising the value of \( \text{K}_w \). Consequently, the pH of pure water decreases below 7 as temperature increases. However, the water is still neutral because \( [\text{H}^+] \) still exactly equals \( [\text{OH}^-] \). Neutral does not mean pH 7; it means \( [\text{H}^+] = [\text{OH}^-] \)!
pH of Strong Bases
Strong bases, like sodium hydroxide (\( \text{NaOH} \)) or potassium hydroxide (\( \text{KOH} \)), dissociate completely in water to release hydroxide ions:
\( \text{NaOH(aq)} \rightarrow \text{Na}^+\text{(aq)} + \text{OH}^-\text{(aq)} \)
To find the pH of a strong base solution, we use a two-step process:
- Find \( [\text{OH}^-] \) from the concentration of the base. For monoprotic bases like \( \text{NaOH} \), \( [\text{OH}^-] = [\text{Base}] \). For diprotic bases like barium hydroxide, \( \text{Ba(OH)}_2 \), \( [\text{OH}^-] = 2 \times [\text{Base}] \).
- Use \( \text{K}_w \) to calculate \( [\text{H}^+] \):
\( [\text{H}^+] = \frac{\text{K}_w}{[\text{OH}^-]} \)
- Calculate the pH of the solution using \( \text{pH} = -\log_{10}[\text{H}^+] \).
Step 1: Determine the hydrogen ion concentration:
Sulfuric acid is a strong diprotic acid. Each molecule releases two protons:
\( [\text{H}^+] = 2 \times 0.040 = 0.080\text{ mol dm}^{-3} \)
Step 2: Calculate the pH:
\[ \text{pH} = -\log_{10}(0.080) = 1.10 \]
The pH of the solution is 1.10 (to 2 decimal places).
Step 1: Determine the hydroxide ion concentration:
Sodium hydroxide is a strong monoprotic base, so:
\( [\text{OH}^-] = 0.025\text{ mol dm}^{-3} \)
Step 2: Calculate the hydrogen ion concentration using Kw:
\[ [\text{H}^+] = \frac{\text{K}_w}{[\text{OH}^-]} = \frac{1.00 \times 10^{-14}}{0.025} = 4.00 \times 10^{-13}\text{ mol dm}^{-3} \]
Step 3: Calculate the pH:
\[ \text{pH} = -\log_{10}(4.00 \times 10^{-13}) = 12.40 \]
The pH of the strong base solution is 12.40 (to 2 decimal places).
Step 1: Calculate the hydrogen ion concentration in pure water:
In pure water, \( [\text{H}^+] = [\text{OH}^-] \), so:
\[ [\text{H}^+] = \sqrt{\text{K}_w} = \sqrt{4.02 \times 10^{-14}} = 2.005 \times 10^{-7}\text{ mol dm}^{-3} \]
Step 2: Calculate the pH:
\[ \text{pH} = -\log_{10}(2.005 \times 10^{-7}) = 6.70 \]
Step 3: State the acid-base nature:
Even though the pH is 6.70 (which is below 7.00), the water is neutral. This is because the concentration of hydrogen ions still exactly equals the concentration of hydroxide ions (\( [\text{H}^+] = [\text{OH}^-] \)).
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