7.3.5. Calculations involving pH

For strong acids, the concentration of H$^+$ is directly equal to the initial concentration of the acid.

When calculating pH for strong acids, we can make a straightforward calculation because strong acids completely dissociate in water. This means that every molecule of the acid donates a hydrogen ion (H$^+$) to the solution. If you start with a strong acid of a certain concentration, that concentration is the same as the concentration of H$^+$ ions in the solution. For example, if you have a 0.1 M solution of hydrochloric acid (HCl), it will fully dissociate to produce 0.1 M of H$^+$.

Think of a glass filled with ice cubes (representing the acid). When the ice melts completely (indicating dissociation), the water level rises to exactly the same height (the concentration of H$^+$) as the original volume of ice cubes (the initial concentration of the acid).

For strong bases, the concentration of OH$^-$ is directly equal to the initial concentration of the base (for monoprotic bases like NaOH). You can then use the pH + pOH = 14.00 relationship to find the pH.

Similar to strong acids, for strong bases, the concentration of hydroxide ions (OH$^-$) produced in solution is equal to the initial concentration of the base since they fully dissociate in water. For example, in a 0.1 M sodium hydroxide (NaOH) solution, you will get 0.1 M of OH$^-$ ions. To find the pH, you first calculate the pOH by using the formula: pOH = -log[OH$^-$]. After calculating the pOH, you can then find the pH using the relationship pH + pOH = 14.00.

Imagine a sponge soaking up water. The more water you pour in (the initial concentration of the base), the more fully soaked the sponge becomes, corresponding to the hydroxide ions being produced. Then, to find the acidity of the solution (pH), you check how much water is absorbed and relate it to the total amount (14) available in the solution.

Since weak acids and bases only partially dissociate, calculating their pH requires using their respective equilibrium dissociation constants (K$_a$ or K$_b$). An ICE (Initial, Change, Equilibrium) table is typically employed to determine the equilibrium concentrations of H$^+$ or OH$^-$.

Weak acids and bases do not completely dissociate in solution. Therefore, calculating their pH requires a more detailed approach than what is used for strong acids and bases. We typically create an ICE (Initial, Change, Equilibrium) table to help visualize the concentrations at various points of the reaction. The equilibrium constant (K$_a$ for acids and K$_b$ for bases) describes how far the reaction proceeds towards products and is essential for determining the concentrations of the free H$^+$ or OH$^-$. From these concentrations, the pH can be calculated as you normally would using pH = -log[H$^+].

Think of a half-filled bottle of soda that you shake. When opened, some bubbles escape immediately (the ions that dissociate), but not all. To figure out how much fizz remains, you’d need to keep track of how many bubbles are created and how many are left (using the ICE table and the K$_a$ or K$_b$ values).

For very weak acids/bases, or when the initial concentration is high, approximations can sometimes be made if the extent of dissociation is very small (less than 5%). Otherwise, the quadratic formula may be necessary to solve for the equilibrium concentrations.

In dealing with weak acids and bases, if the concentration is relatively high, i.e., greater than or equal to 0.1 M, it is sometimes acceptable to simplify the calculations by assuming that the change in concentration due to dissociation is negligible (less than 5%). However, for situations where the dissociation isn't negligible, or when the acid/base is very weak, you would have to set up a quadratic equation based on the equilibrium expression to solve for the concentrations at equilibrium.

Imagine you are filling a basin with a small leak. If the leak is tiny relative to the volume of water you're adding, you can ignore it and assume the basin is full. However, if the leak is large relative to the water volume, you must account for it, which might require more complex calculations to figure out how much water is left.

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