Question

Consider \(50.0 \mathrm{~mL}\) of a solution of weak acid \(\mathrm{HA}=K_{\mathrm{a}}(1.00 \times\) \(10^{-6}\) ), which has a pH of \(4.000\). What volume of water must be added to make the \(\mathrm{pH}=5.000 ?\)

Short answer

To increase the pH of the 50.0 mL solution of weak acid HA from 4.000 to 5.000, 50 mL of water must be added.

Step-by-step solution

Calculate the initial concentration of HA

First, we need to calculate the concentration of the weak acid HA before adding water. We can use the pH formula to do this. pH = -log10[H+] We know the initial pH is 4.000, so we can calculate the initial concentration of H+ ions: 4.000 = -log10[H+] [H+] = 10^(-4.000) Now, we can find the concentration of HA by using Ka: Ka = [H+][A-] / [HA] Since the acid is weak, the concentration of A- and H+ will be approximately equal. Thus, we can simplify the equation as follows: 1.00 × 10^(-6) = [H+]² / [HA] [HA] = [H+]² / (1.00 × 10^(-6)) Substitute the value of [H+] in the equation: [HA] = (10^(-4.000))² / (1.00 × 10^(-6)) Calculate the initial concentration of HA using the given values: [HA] = (10^(-8))/(1.00 × 10^(-6)) = 0.01 M

Calculate the moles of HA

Next, we need to find the moles of HA in the 50.0 mL solution. We can use the following formula: moles = concentration × volume Substitute the values for concentration and volume: moles = 0.01 M × 50.0 mL = 0.01 M × 0.05 L = 0.0005 moles

Calculate the final concentration of HA

After adding water to the solution, we need to determine the final concentration of HA in the new solution. We can use the formula: concentration = moles / volume Since we don't know the final volume yet, we can represent the volume of water added as 'x.' Thus, the final volume of the solution is (0.05 + x) L. Substitute the values into the formula: [HA]_final = 0.0005 moles / (0.05 + x) L

Calculate the final concentration of H+

Now, we need to find the final concentration of H+ ions, using the given final pH: 5.000 = -log10[H+] [H+]_final = 10^(-5.000)

Determine the volume of water to be added

We'll use the final concentration of HA, Ka, and the final concentration of H+ to find the final volume: Ka = [H+]_final[A-]_final / [HA]_final As the final concentration of A- is approximately equal to [H+]_final, we can simplify the equation as follows: 1.00 × 10^(-6) = 10^(-10) / [HA]_final Substitute the [HA]_final expression from Step 3: 1.00 × 10^(-6) = 10^(-10) / (0.0005 moles / (0.05 + x) L) Rearrange and solve for x: (0.0005 moles / (1.00 × 10^(-6))) = (0.05 + x) * 10^(-10) 0.05 + x = (0.0005 * 10^4) x = (0.0005 * 10^4) - 0.05 Calculate the value of x: x = 0.05 L = 50 mL

Conclusion

To increase the pH of the 50.0 mL solution of weak acid HA from 4.000 to 5.000, 50 mL of water must be added.

Key concepts

Acid Dissociation Constant (Ka)

Understanding the acid dissociation constant, commonly represented as Ka, is crucial when dealing with weak acids and pH calculations. Ka is a quantitative measure of an acid's strength – the higher the Ka value, the stronger the acid and the more it dissociates in solution. Ka is determined based on the equilibrium concentrations of the acid (HA), the conjugate base (A-), and the hydrogen ions (H+) in the solution.

For a weak acid dissociating according to the equation HA ⇌ H+ + A-, the Ka is given as: \[Ka = \frac{[H^+][A^-]}{[HA]}\]
In simpler terms, it indicates how much of the acid has dissociated to form hydrogen ions and its conjugate base in solution at equilibrium. In our exercise, we're given that the Ka value of the weak acid is \(1.00 \times 10^{-6}\), representing its relatively low tendency to dissociate.

pH and pOH Calculations

pH is a scale used to specify the acidity or basicity of an aqueous solution. It is calculated as the negative base-10 logarithm of the hydrogen ion concentration: \[pH = -\log_{10} [H^+]\]
Similarly, pOH is related to the concentration of hydroxide ions (OH-) and is calculated as: \[pOH = -\log_{10} [OH^-]\]
The pH and pOH scales are interconnected by the relationship pH + pOH = 14, which holds true at 25°C. In the context of this exercise, we use the pH value to calculate the hydrogen ion concentration, vital information for understanding the acidity of the solution. A pH of 4 indicates a more acidic solution compared to a pH of 5, corresponding to a tenfold decrease in the hydrogen ion concentration.

Molarity and Dilution

Molarity (M) is a measure of the concentration of a solute in a solution and is defined as the number of moles of solute divided by the volume of solution in liters. The formula for molarity is: \[M = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\]
Dilution involves adding solvent to a solution, which decreases the solute concentration while keeping the total number of moles constant. The dilution formula, \(M_1V_1 = M_2V_2\), where M1 and V1 are the molarity and volume before dilution, and M2 and V2 after dilution, is an essential concept for adjusting the pH of solutions, as demonstrated in our problem. To achieve the desired pH, we add water (dilute the solution) and calculate the new volume while ensuring the number of moles of acid remains constant.

Hydrogen Ion Concentration

The hydrogen ion concentration, [H+], is a direct measure of the acidity in a solution. For any aqueous solution, the concentration of hydrogen ions determines its pH value. In every step of our pH calculation, we refer back to the hydrogen ion concentration as we convert between pH values and [H+], using the mathematical relationship: \[pH = -\log_{10} [H^+]\]
In the case of our exercise, the initial and final [H+] values guide us in calculating the volumes required for dilution. By adjusting the hydrogen ion concentration, or indirectly the pH value, we can tailor the acidity of a solution to fit specific needs in various scientific applications.