Complete the table below: Note: Make simplifying assumptions, do not use the quadratic formula.
What is the pH of the solution created by combining 2.00 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)?
Complete the table below: Note: Make simplifying assumptions, do not use the quadratic formula.
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)?
See your post above.
To determine the pH of a solution, we need to consider the concentration of hydrogen ions (H+) in the solution. The pH is calculated using the formula:
pH = -log[H+]
To find the concentration of H+ in the solution, we need to use the concept of stoichiometry and apply the principle of conservation of mass.
In the first case, combining 2.00 mL of 0.10 M NaOH(aq) with 8.00 mL of 0.10 M HCl(aq), we can assume that the reaction between NaOH and HCl is complete. Therefore, we can determine the concentration of H+ by using stoichiometric ratios.
1 mol of NaOH reacts with 1 mol of HCl, so the concentration of H+ in the final solution can be calculated as follows:
Initial moles of NaOH = (2.00 mL)(0.10 mol/L)(1 L / 1000 mL) = 0.0002 mol
Initial moles of HCl = (8.00 mL)(0.10 mol/L)(1 L / 1000 mL) = 0.0008 mol
According to the stoichiometry, the number of moles of HCl that will react with NaOH is equal to the number of moles of NaOH added. Therefore, the concentration of H+ in the final solution is:
Final moles of H+ = 0.0002 mol
Since the total volume of the solution is the sum of the volumes of NaOH and HCl, we add the volumes:
Total volume of solution = 2.00 mL + 8.00 mL = 10.00 mL = 0.010 L
Concentration of H+ = Final moles of H+ / Total volume of solution
Concentration of H+ = 0.0002 mol / 0.010 L = 0.020 M
Now that we have the concentration of H+, we can calculate the pH using the formula:
pH = -log[H+]
pH = -log(0.020)
pH ≈ 1.70
Therefore, the pH of the solution created by combining 2.00 mL of 0.10 M NaOH(aq) with 8.00 mL of 0.10 M HCl(aq) is approximately 1.70.
For the second case, when 8.00 mL of 0.10 M HC2H3O2(aq) is diluted with 100 mL of water, we need to consider the dilution factor.
The dilution factor is calculated by dividing the final volume of the solution by the initial volume:
Dilution factor = Final volume / Initial volume
Dilution factor = (8.00 mL + 100 mL) / 8.00 mL
Dilution factor = 12.50
Since the dilution factor is 12.50, the concentration of HC2H3O2 in the final solution is:
Initial concentration of HC2H3O2 = 0.10 M
Final concentration of HC2H3O2 = 0.10 M / 12.50
Final concentration of HC2H3O2 = 0.008 M
Now, we can calculate the concentration of H+ using stoichiometry.
The dissociation of HC2H3O2 can be represented by the following equilibrium equation:
HC2H3O2(aq) ⇌ H+(aq) + C2H3O2-(aq)
Since HC2H3O2 is a weak acid, we need to consider the initial concentration of the acid and the dissociation constant (Ka) to find the concentration of H+.
Ka is the equilibrium constant for the dissociation of HC2H3O2 and is equal to:
Ka = [H+][C2H3O2-] / [HC2H3O2]
We can assume that the concentration of H+ at equilibrium is the same as the initial concentration of H+. Therefore, we can approximately consider the concentration of H+ as the concentration of HC2H3O2.
H+ concentration = 0.008 M
The pH can be calculated using the formula:
pH = -log[H+]
pH = -log(0.008)
pH ≈ 2.10
Therefore, the pH of the solution created by combining 2.00 mL of 0.10 M NaOH(aq) with 8.00 mL of 0.10 M HC2H3O2(aq), after diluting the HC2H3O2 with 100 mL of water, is approximately 2.10.