Question

An acid with a \(\mathrm{p} K_{\mathrm{a}}\) of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the protonated to the deprotonated form of the acid?\( {6}\)

Short answer

The ratio of protonated to deprotonated form is 100.

Step-by-step solution

Understand the Problem

We're given the \( \text{pH} \) of the solution as 6.0 and the \( \text{p}K_a \) of the acid as 8.0. We need to find the ratio of the protonated form \([\text{HA}]\) to the deprotonated form \([\text{A}^-]\).

Use the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is \( \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \).

Rearrange the Equation

Rearrange the equation to solve for the ratio \( \frac{[\text{A}^-]}{[\text{HA}]} \) as follows:\[\log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) = \text{pH} - \text{p}K_a\]

Substitute Known Values

Substitute \( \text{pH} = 6.0 \) and \( \text{p}K_a = 8.0 \) into the equation:\[\log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) = 6.0 - 8.0 = -2.0\]

Calculate the Ratio

Use the property of logarithms to find the ratio:\[\frac{[\text{A}^-]}{[\text{HA}]} = 10^{-2} = 0.01\]

Find the Ratio of Protonated to Deprotonated

The question asks for the ratio of \([\text{HA}]\) to \([\text{A}^-]\), which is the inverse of what we have calculated:\[\frac{[\text{HA}]}{[\text{A}^-]} = \frac{1}{0.01} = 100\]

Key concepts

pH calculation

pH calculation is a fundamental concept in chemistry that helps us understand the acidity or basicity of a solution. The scale ranges from 0 to 14:
- Values below 7 indicate acidic solutions. - Values above 7 indicate basic solutions. - A pH of 7 is neutral, like pure water.To calculate pH, one generally uses the formula \[\text{pH} = -\log[\text{H}^+]\]where \([\text{H}^+]\) is the concentration of hydrogen ions. This formula shows how pH is a logarithmic measure, meaning each whole number change in pH represents a tenfold change in hydrogen ion concentration. This results in a very sensitive scale, where small differences in pH are significant.In the context of acid-base chemistry, understanding pH can help us predict how acids and bases behave in a solution. In our case, knowing the pH of a solution allows us to use the Henderson-Hasselbalch equation to find the ratio of different species in acid-base equilibrium.

acid-base equilibrium

Acid-base equilibrium is a chemical balance established between an acid and its corresponding base in a solution. When an acid dissolves in water, it donates protons (H⁺ ions) and establishes an equilibrium between the:- Undissociated (protonated) acid molecules - Dissociated (deprotonated) base ions.
An important measure associated with this equilibrium is the \(\text{p}K_a\), which measures the strength of an acid. The \(\text{p}K_a\) value helps us understand how likely the acid is to donate a proton. Lower \(\text{p}K_a\) values indicate stronger acids that donate protons more readily, while higher \(\text{p}K_a\) values indicate weaker acids. The Henderson-Hasselbalch equation, \(\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\), ties these concepts together. It accounts for the relationship between pH, \(\text{p}K_a,\)and the concentrations of the protonated and deprotonated forms in solution, helping us understand and predict how the components in an acid-base equilibrium behave in different pH environments.

protonated vs deprotonated ratio

The protonated versus deprotonated ratio tells us about the distribution of an acid's two forms in solution:- The protonated form \([\text{HA}]\)contains the H⁺ ion, remaining as an acid molecule.- The deprotonated form \([\text{A}^-]\)results when the acid loses a proton, leaving a base ion.Using the Henderson-Hasselbalch equation, we can find this ratio by knowing the \(\text{pH}\)and \(\text{p}K_a\) of the solution. The equation \(\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\)can be rearranged to show the ratio:\[\log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = \text{pH} - \text{p}K_a\]This lets you calculate the deprotonated to protonated ratio. To find the protonated versus deprotonated ratio, simply take the inverse.For example, if \(\log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = -2\), then \(\frac{[\text{A}^-]}{[\text{HA}]}\)=0.01, so \(\frac{[\text{HA}]}{[\text{A}^-]}\)=100. This shows the solution has far more protonated molecules than deprotonated ones, reflecting the equilibrium point given by the solution's specific \(\text{pH}\)and \(\text{p}K_a\).