Problem 112 - Chemistry Matter and Change

Question

You have 20.0 \(\mathrm{mL}\) of a solution of a weak acid, \(\mathrm{HX}\) , whose \(K_{\mathrm{a}}\) equals \(2.14 \times 10^{-6} .\) The \(\mathrm{pH}\) of the solution is found to be \(3.800 .\) How much distilled water would you have to add to the solution to increase the pH to 4.000\(?\)

Step-by-Step Solution

Answer

Add 11.6 mL of distilled water.

Step 1: Understanding the Initial pH

The initial pH is given as 3.800, which we use to determine the initial concentration of \( ext{H}^+\). Recall that \( ext{pH} = -\log[ ext{H}^+]\). Calculating, we find \([ ext{H}^+] = 10^{-3.800} \approx 1.58 \times 10^{-4} \text{M}\).

Step 2: Calculate Initial \\([ ext{HX}])\\

Using the weak acid dissociation constant, \(K_a\), set up the expression for \(K_a\) given by \(K_a = \frac{[ ext{H}^+][ ext{X}^-]}{[ ext{HX}]}\). With \(K_a = 2.14 \times 10^{-6}\) and \([ ext{H}^+] = [ ext{X}^-] = 1.58 \times 10^{-4} \text{M}\), solve for \([ ext{HX}]\). \(K_a = \frac{(1.58 \times 10^{-4})^2}{[ ext{HX}]},\) giving \([ ext{HX}] \approx 0.0117 \text{M}\).

Step 3: Determine Desired [H⁺] for pH 4.000

For a pH of 4.000, the concentration of \([ ext{H}^+]\) is \(10^{-4.000} = 1.00 \times 10^{-4} \text{M}\).

Step 4: Use Dilution Formula to Find Final Volume

Use the dilution relationship \(M_1V_1 = M_2V_2\), where \(M_1 = 1.58 \times 10^{-4} \text{M}, V_1 = 20.0 \text{mL}, M_2 = 1.00 \times 10^{-4} \text{M}\). Solve for \(V_2\) using the equation: \(1.58 \times 10^{-4} \text{M} \times 20.0 \text{mL} = 1.00 \times 10^{-4} \text{M} \times V_2\), which gives \(V_2 = 31.6 \text{mL}\).

Step 5: Calculate the Volume of Water to Add

The total volume after dilution to achieve the desired pH is 31.6 mL. Subtract the original solution volume from this: \(V_ ext{water} = 31.6 \text{mL} - 20.0 \text{mL} = 11.6 \text{mL}\).

Key Concepts

Weak Acid Dissociation

When we talk about weak acid dissociation, we're discussing how a weak acid like \(\text{HX}\) partially breaks apart in water to form ions. Unlike strong acids, which dissociate completely, weak acids such as \(\text{HX}\) only partially dissociate into \(\text{H}^+\) ions and its conjugate base \(\text{X}^-\). This dissociation is reversible and is characterized by the acid's dissociation constant, \(K_a\).

The equilibrium expression for this can be given as:

• \(K_a = \frac{[\text{H}^+][\text{X}^-]}{[\text{HX}]}\)

Here, \([\text{H}^+]\) and \([\text{X}^-]\) are the molar concentrations of the ions produced, and \([\text{HX}]\) is the concentration of the non-dissociated acid.

Knowing \(K_a\) helps us predict how much the acid will dissociate at a given concentration. For example, with a \(K_a\) of \(2.14 \times 10^{-6}\), as given in our problem, we can infer that \(\text{HX}\) is a weak acid, meaning it doesn't release many \(\text{H}^+\) ions. This minimal ion production directly affects solutions' electrical conductivity and also occupies a unique niche in buffer solutions, where weak acids and their conjugates resist changes in pH.

pH Calculation

To talk pH, we first need to understand it's a way to measure how acidic or basic a solution is. The formula is simple, but powerful:

• \(\text{pH} = -\log[\text{H}^+]\)

The pH scale ranges from 0 to 14:

• Lower pH means more acidic (more \(\text{H}^+\))
• Higher pH means more basic (more \(\text{OH}^-\))

The initial pH given for the solution in the problem was 3.800. That indicates the solution is acidic and, based on the calculations, the concentration of \(\text{H}^+\) ions was approximately \(1.58 \times 10^{-4} \text{M}\). By adjusting the pH to 4.000, we change the concentration of \(\text{H}^+\) ions to \(1.00 \times 10^{-4} \text{M}\), a slight shift towards a less acidic solution. This demonstrates how pH is not a linear scale; each whole number change represents a tenfold change in hydrogen ion concentration.

Dilution Formula

Dilution is essentially the process of adding more solvent (like water), which decreases the concentration of solutes (like \(\text{H}^+\) ions) in the solution. It's extremely handy in experiments like this one, where we need to adjust the pH by changing concentration.

We apply the **Dilution Formula**:

• \(M_1V_1 = M_2V_2\)

Here:

• \(M_1\) and \(V_1\) are the molarity and volume of the starting solution
• \(M_2\) and \(V_2\) are the same for the diluted solution

Given our problem's specifics, we use the formula to solve for the final volume \(V_2\) needed to reach the desired pH of 4.000. By substituting in the known values—\(M_1 = 1.58 \times 10^{-4} \text{M}\), and \(M_2 = 1.00 \times 10^{-4} \text{M}\) while letting \(V_1\) be the initial 20 mL solution volume—we find \(V_2\) to be about 31.6 mL. So, the addition of 11.6 mL of water achieves this, giving us a larger volume of a less concentrated, slightly less acidic solution. This step is vital in chemical analysis and preparatory labs for manipulating solution strengths.