Question

Calculate the concentration of all species present and the pH of a $0.020-M$ HF solution.

Short answer

The concentration of all species present in the 0.020-M HF solution are approximately: [HF] = 0.020-x M, [H+] = 0.0026 M, [F-] = 0.0026 M, and the pH of the solution is approximately 2.59.

Step-by-step solution

Write the reaction for the dissociation of the weak acid HF

In this step, write the balanced equation for the dissociation of the weak acid, HF, into its ions: $$HF(aq)\leftrightarrow H^{+}(aq)+F^{-}(aq)$$

Write the Ka expression for HF

Write the acid dissociation constant (Ka) expression for HF dissociation: $$K_a=\frac{[H^{+}][F^{-}]}{[HF]}$$

Create an equilibrium table

Express the initial concentrations, changes, and equilibrium concentrations as a table using the initial concentration of HF (0.020 M) and variables x: | | HF | H+ | F- | |:---:|:-----:|:--:|:--:| |Initial|0.020 M| 0 | 0 | |Change |-x | +x | +x | |Equilibrium|0.020-x| x | x |

Substitute the variables in the Ka expression and solve for x

Substitute the variables of the equilibrium expressions into the Ka expression and use the given Ka value for HF(6.8 × 10^(-4)). $$(6.8\times {10}^{-4})=\frac{x^2}{0.020-x}$$ Assume x is very small compared to 0.020, hence we can ignore the x in the denominator. $$(6.8\times {10}^{-4})\approx \frac{x^2}{0.020}$$ Now, solve for x (H+ concentration): $$x\approx \sqrt{(6.8\times {10}^{-4})(0.020)}$$ $$x\approx 0.0026\text{M}$$

Calculate the pH

Use the concentration of H+ (x) to calculate the pH: $$pH=-\log [H^{+}]$$ $$pH\approx -\log (0.0026)$$ $$pH\approx 2.59$$ Therefore, the concentration of all species present are approximately: [HF] = 0.020-x M, [H+] = 0.0026 M, [F-] = 0.0026 M, and the pH of the 0.020 M HF solution is approximately 2.59.

Key concepts

Acid Dissociation Constant (Ka)

The Acid Dissociation Constant, represented as $K_a$, is a crucial parameter in understanding the behavior of weak acids in an aqueous solution. It quantifies the extent to which an acid dissociates into its ions. A larger $K_a$ value indicates a stronger acid that dissociates more completely, whereas a smaller $K_a$ value points to a weaker acid that remains largely undissociated.

For hydrofluoric acid (HF), the dissociation process is expressed as follows:
- HF dissociates into $H^{+}$ ions and $F^{-}$ ions
- The reaction is reversible, shown as: $$HF\rightleftharpoons H^{+}+F^{-}$$

The equilibrium expression for HF's dissociation can be described as:
$$K_a=\frac{[H^{+}][F^{-}]}{[HF]}$$

Understanding this concept allows us to determine how much of the acid dissociates and calculate the concentrations of each species present at equilibrium.

pH Calculation

Calculating the pH of an acidic solution is essential for understanding the solution's acidity level. The pH is determined by the concentration of hydrogen ions $[H^{+}]$ in the solution. It is calculated using the formula:

$$pH=-\log [H^{+}]$$

For a weak acid like HF, after calculating the $[H^{+}]$ concentration through the equilibrium process, we can find the pH. In this case, the $[H^{+}]$ was found to be approximately 0.0026 M.

Let's use this concentration to find the pH:
- Substitute $[H^{+}]=0.0026M$ into the pH formula.
- The calculation becomes: $$pH=-\log (0.0026)$$
- Therefore, the pH of the solution is approximately 2.59.

A low pH value, like 2.59, signifies a relatively high concentration of hydrogen ions, indicating the solution is acidic.

Equilibrium Concentration

In an acid-base reaction, the concept of equilibrium concentration reveals how the concentrations of reactants and products adjust until the system stabilizes. To calculate these concentrations, we use an equilibrium table often called an ICE table, standing for Initial, Change, and Equilibrium.

For the HF solution:

• Initial: Begin with an initial concentration of 0.020 M for HF.
  
• Change: As dissociation occurs, $x$ amount of HF converts into $H^{+}$ and $F^{-}$.
• Equilibrium: The concentrations become $[HF]=0.020-x$, $[H^{+}]=x$, $[F^{-}]=x$.

The equilibrium concentration is essential because it guides us in calculating the pH and other properties of the solution.

By approximating that $x$ is small compared to the initial concentration, we simplify the math to: $[HF]=0.020$, $[H^{+}]=0.0026$, $[F^{-}]=0.0026$. This simplifies solving the quadratic equation usually involved in such calculations.

HF Dissociation

HF, or hydrofluoric acid, is a weak acid with a specific behavior in aqueous solutions. Its dissociation in water is a typical example of weak acid equilibria.

The balanced equation for HF dissociation is:
- $$HF\rightleftharpoons H^{+}+F^{-}$$

This equation shows the weak acid dissociating into protons $H^{+}$ and fluoride ions $F^{-}$ in water. Notably, only a small fraction of HF molecules undergo dissociation, unlike strong acids that dissociate completely.

The dissociation process can be described quantitatively by using its $K_a$ value. For HF, with a $K_a$ of $6.8\times {10}^{-4}$, we see it's not a large number, reflecting its weak acidic nature.

Understanding HF dissociation is vital for calculating the concentrations of all involved species at equilibrium and the pH of the solution. It shows how equilibrium adjusts as the system reaches a steady state where the rate of dissociation equals the rate of recombination.