What is the pH of the solution created by combining 2.70 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)?
What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water (like it will be in the experiment you perform in lab)?
they ask you to find the pH for both the HCl and the HC2H3O2.
i understand the first question its just the second question that is confusing me
The part about dilution to 100.
For the NaOH/HCl part, it will be (H^+) = moles excess HCl/final volume in liters and pH from that.
For the buffer part (HAc/Ac^-), the dilution has no effect because with the Henderson-Hasselbalch equation you are diluting both the base and the acid th same amount and the ratio doesn't change.
pH = pKa + log (base)/(acid)
To find the pH of a solution, we need to know the concentration of hydrogen ions (H+) in the solution. The concentration of H+ ions can be calculated using the concept of molarity (M) and the equation for the ionization of an acid.
For the first question, we are given the volumes and concentrations of NaOH and HCl. Since NaOH is a strong base and HCl is a strong acid, their reaction will result in the formation of water and a neutral solution. Therefore, the pH of this solution is 7 (neutral pH).
Now, let's focus on the second question, where we need to consider the dilution of the acid solution. Dilution is a process that involves adding solvent (usually water) to reduce the concentration of the solute.
In this case, the 8.00 mL of 0.10 M acid is first diluted with 100 mL of water. To determine the new concentration of the acid, we need to apply the dilution formula, which states that the initial concentration multiplied by the initial volume equals the final concentration multiplied by the final volume.
Initial concentration x initial volume = final concentration x final volume
(0.10 M) x (8.00 mL) = final concentration x (100 mL + 8.00 mL)
0.100 M x 8.00 mL = final concentration x 108.00 mL
Now we solve for the final concentration:
final concentration = (0.100 M x 8.00 mL) / 108.00 mL
final concentration = 0.007407 M
Now that we have the final concentration of the acid after dilution, we can calculate the pH.
For HCl:
Since HCl is a strong acid, it completely ionizes in water and produces H+ ions. Therefore, the concentration of H+ ions is equal to the concentration of the acid (0.007407 M). To find the pH, we can use the formula:
pH = -log[H+]
pH = -log(0.007407)
pH = 2.13
For HC2H3O2 (acetic acid):
Acetic acid is a weak acid, and its ionization follows a reversible reaction:
HC2H3O2 ⇌ H+ + C2H3O2-
To determine the concentration of the H+ ions, we need to consider the equilibrium constant (Ka) for acetic acid, which is 1.8 x 10^-5 at 25°C. Using an ICE table (Initial, Change, Equilibrium), we can calculate the concentration of H+ ions at equilibrium.
Let's assume x mol/L is the concentration of H+ ions in equilibrium. Then, the concentration of HC2H3O2 would be (0.007407 - x) mol/L, and the concentration of C2H3O2- would also be x mol/L.
Applying the equilibrium expression for acetic acid (Ka = [H+][C2H3O2-] / [HC2H3O2]), we get:
1.8 x 10^-5 = (x)(x) / (0.007407 - x)
Simplifying the equation, we can approximate that x is small compared to 0.007407. Thus, we can assume that 0.007407 - x ≈ 0.007407.
1.8 x 10^-5 ≈ x^2 / 0.007407
x^2 = 1.8 x 10^-5 x 0.007407
x^2 ≈ 1.333 x 10^-7
Taking the square root of both sides, we find:
x ≈ √(1.333 x 10^-7)
x ≈ 3.65 x 10^-4 mol/L
Since we assumed the initial concentration of acid, HC2H3O2, to be 0.007407 M, we should subtract the concentration of H+ ions from the initial acid concentration to get the concentration of unreacted HC2H3O2:
concentration of unreacted HC2H3O2 = 0.007407 M - 3.65 x 10^-4 M
concentration of unreacted HC2H3O2 ≈ 0.007041 M
Now that we have the concentrations of H+ and HC2H3O2, we can calculate the pH using the equation:
pH = -log[H+]
pH = -log(3.65 x 10^-4)
pH ≈ 3.44
Therefore, the pH of the solution after dilution for HCl is 2.13, and for HC2H3O2 it is approximately 3.44.