Question

Calculate the $\mathrm{pH}$ of an aqueous solution containing $1.0\times {10}^{-2}M$ $\mathrm{HCl},1.0\times {10}^{-2}\mathrm{M}{\mathrm{H}}_2{\mathrm{SO}}_4$, and $1.0\times {10}^{-2}\mathrm{M}\mathrm{HCN}$.

Short answer

The pH of the aqueous solution containing $1.0\times {10}^{-2}M$ $\mathrm{HCl}$, $1.0\times {10}^{-2}\mathrm{M}{\mathrm{H}}_2{\mathrm{SO}}_4$, and $1.0\times {10}^{-2}\mathrm{M}\mathrm{HCN}$ is approximately 1.70.

Step-by-step solution

Calculate ${\mathrm{H}}^{+}$ concentration from $\mathrm{HCl}$

Since $\mathrm{HCl}$ is a strong acid, we assume it fully dissociates in water: $\mathrm{HCl}\to {\mathrm{H}}^{+}+{\mathrm{Cl}}^{-}$ The initial concentration of $\mathrm{HCl}$ is $1.0\times {10}^{-2}M$. As it fully dissociates, the ${\mathrm{H}}^{+}$ concentration coming from $\mathrm{HCl}$ is the same as the initial concentration of $\mathrm{HCl}$: $$[{\mathrm{H}}^{+}{]}_{\mathrm{HCl}}=1.0\times {10}^{-2}M$$

Calculate ${\mathrm{H}}^{+}$ concentration from ${\mathrm{H}}_2{\mathrm{SO}}_4$

For the first dissociation of ${\mathrm{H}}_2{\mathrm{SO}}_4$, which is a strong acid, it dissociates as follows: ${\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{H}}^{+}+{\mathrm{HSO}}_4^{-}$ The initial concentration of ${\mathrm{H}}_2{\mathrm{SO}}_4$ is $1.0\times {10}^{-2}M$. As only the first dissociation is strong, the ${\mathrm{H}}^{+}$ concentration coming from ${\mathrm{H}}_2{\mathrm{SO}}_4$ is also the same as the initial concentration of ${\mathrm{H}}_2{\mathrm{SO}}_4$: $$[{\mathrm{H}}^{+}{]}_{{\mathrm{H}}_2{\mathrm{SO}}_4}=1.0\times {10}^{-2}M$$

Calculate ${\mathrm{H}}^{+}$ concentration from $\mathrm{HCN}$

Now for $\mathrm{HCN}$, a weak acid, we must use its $K_a$ value to find the ${\mathrm{H}}^{+}$ contribution. The acid dissociation for $\mathrm{HCN}$ is: $\mathrm{HCN}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{CN}}^{-}$ For this equation, the $K_a$ expression is: $$K_a=\frac{[{\mathrm{H}}^{+}][{\mathrm{CN}}^{-}]}{[\mathrm{HCN}]}$$ Given $K_a=6.2\times {10}^{-10}$ for $\mathrm{HCN}$ and the initial concentration, $1.0\times {10}^{-2}M$, we can set up an ICE (Initial, Change, Equilibrium) table: | | ${\mathrm{H}}^{+}$ | ${\mathrm{CN}}^{-}$ | $\mathrm{HCN}$ | |---------|--------------|----------------|-------------| | Initial | $x$ | $0$ | $1.0\times {10}^{-2}$ | | Change | +$x$ | +$x$ | -$x$ | | Eqm | $x$ | $x$ | $1.0\times {10}^{-2}-x$ | Now we can substitute the values into the $K_a$ expression and solve for $x$, which is the ${\mathrm{H}}^{+}$ concentration from $\mathrm{HCN}$: $$6.2\times {10}^{-10}=\frac{x\times x}{1.0\times {10}^{-2}-x}$$ As $K_a$ for $\mathrm{HCN}$ is very small, the dissociation of $\mathrm{HCN}$ is very low, and we can make the assumption that $x\ll 1.0\times {10}^{-2}$. This simplifies the equation to: $$6.2\times {10}^{-10}\approx \frac{x^2}{1.0\times {10}^{-2}}$$ Solving for $x$, we find: $$[{\mathrm{H}}^{+}{]}_{\mathrm{HCN}}\approx 2.5\times {10}^{-6}M$$

Calculate total ${\mathrm{H}}^{+}$ concentration and pH

Now that we have the ${\mathrm{H}}^{+}$ concentrations coming from each acid, we can find the total ${\mathrm{H}}^{+}$ concentration by adding all three contributions: $$[{\mathrm{H}}^{+}{]}_{\text{total}}=[{\mathrm{H}}^{+}{]}_{\mathrm{HCl}}+[{\mathrm{H}}^{+}{]}_{{\mathrm{H}}_2{\mathrm{SO}}_4}+[{\mathrm{H}}^{+}{]}_{\mathrm{HCN}}$$ $$[{\mathrm{H}}^{+}{]}_{\text{total}}=1.0\times {10}^{-2}M+1.0\times {10}^{-2}M+2.5\times {10}^{-6}M\approx 2.0\times {10}^{-2}M$$ Finally, we can use the pH formula to calculate the pH of the solution: $$pH=-{\log }_{10}[{\mathrm{H}}^{+}]$$ $$pH=-{\log }_{10}(2.0\times {10}^{-2}M)\approx 1.70$$ So, the pH of the aqueous solution containing $1.0\times {10}^{-2}M$ $\mathrm{HCl}$, $1.0\times {10}^{-2}\mathrm{M}{\mathrm{H}}_2{\mathrm{SO}}_4$, and $1.0\times {10}^{-2}\mathrm{M}\mathrm{HCN}$ is approximately 1.70.

Key concepts

Strong Acids

Strong acids, such as hydrochloric acid (HCl), are characterized by their ability to completely dissociate in aqueous solutions. This means that when strong acids are added to water, they fully separate into their constituent ions. For instance, when HCl is dissolved in water, it separates into hydrogen ions ( H^+ ) and chloride ions ( Cl^- ).

The complete dissociation implies that the concentration of hydrogen ions ( H^+ ) produced is equal to the initial concentration of the acid. This is why, in the calculation of pH from strong acids like HCl, we can directly use the concentration of the acid as the concentration of hydrogen ions. This property makes strong acids very important in chemical reactions where a large amount of hydrogen ions is needed swiftly.

Weak Acids

Unlike strong acids, weak acids only partially dissociate in water. Hydrocyanic acid (HCN) serves as a typical example. When HCN is dissolved in water, only a small fraction of it splits into hydrogen ions ( H^+ ) and cyanide ions ( CN^- ).

This partial dissociation implies that the concentration of hydrogen ions from a weak acid depends on its acid dissociation constant, denoted as K_a . A weak acid has a smaller K_a value, indicating lesser dissociation. As a result, to calculate the concentration of hydrogen ions from weak acids, you will often need to perform equilibrium calculations using the K_a value and the initial concentration of the acid.

Acid Dissociation

The process by which an acid separates into its ions in a solution is called acid dissociation. This is represented by an equilibrium equation for weak acids and by a complete dissociation for strong acids.

For example, the dissociation of sulfuric acid ( H_2SO_4 ) occurs in steps. In the first dissociation, a hydrogen ion ( H^+ ) is released, and the remainder is the hydrogen sulfate ion ( HSO_4^- ). Sulfuric acid is unique due to its first dissociation being strong and the second one being weaker, adding complexity as only the first dissociation is considered in pH calculations for strong acids.

For weak acids like HCN, a dynamic equilibrium forms between the dissociated and undissociated forms. The effectiveness of the dissociation is quantified by the acid dissociation constant ( K_a ), which helps determine the proportion of ions present under equilibrium.

Concentration of Hydrogen Ions

Understanding the concentration of hydrogen ions is critical in the context of pH calculations. The pH of a solution is a measure of its acidity and is dependent on the concentration of hydrogen ions present.

In a solution with several acids, such as HCl, H_2SO_4, and HCN, each contributes differently to the concentration of hydrogen ions. Strong acids like HCl and the first dissociation of H_2SO_4 provide a straightforward calculation as their hydrogen ion contributions are equivalent to their molar concentrations. Weak acids like HCN require the use of K_a values to determine their contribution.

Once the contributions from all acids are known, they are summed to get the total concentration of hydrogen ions in the solution. The pH can then be calculated using the formula: $pH=-{\log }_{10}[H^{+}]$. By understanding these contributions and the calculations involved, you can predict the pH of any acidic solution accordingly.