calculate the pH and the pOH of an aqueous solution that is 0.035 M In HCI and 0.080 M in HBr?
To calculate the pH and pOH of the aqueous solution, we can use the concentration of HCl and HBr.
First, let's calculate the concentration of H+ ions in the solution:
For HCl:
[H+] = 0.035 M
For HBr:
[H+] = 0.080 M
Now, let's calculate the pH using the formula:
pH = -log[H+]
For HCl:
pH = -log(0.035) = 1.46
For HBr:
pH = -log(0.080) = 1.10
Next, we can calculate the pOH using the formula:
pOH = -log[OH-]
Since we have the concentration of H+ and HCl/HBr are strong acids, we can assume that the concentration of OH- is negligible. Therefore, pOH = 0.
Lastly, we can calculate the pOH using the relationship between pH and pOH:
pH + pOH = 14
For HCl:
pOH = 14 - pH = 14 - 1.46 = 12.54
For HBr:
pOH = 14 - pH = 14 - 1.10 = 12.90
Therefore, the pH and pOH of the solution containing 0.035 M HCl and 0.080 M HBr are:
pH = 1.46 (for HCl)
pH = 1.10 (for HBr)
pOH = 12.54 (for HCl)
pOH = 12.90 (for HBr)
To calculate the pH and pOH of the given aqueous solution, we need to use the concept of dissociation of acids in water and the relationship between pH, pOH, and the concentration of hydrogen ions (H+) and hydroxide ions (OH-) in a solution.
First, we need to determine the concentration of H+ ions in the solution. In this case, we have two acids, HCl and HBr.
HCl dissociates in water as follows:
HCl(aq) → H+(aq) + Cl-(aq)
HBr dissociates in water as follows:
HBr(aq) → H+(aq) + Br-(aq)
Since both HCl and HBr dissociate completely, the concentration of H+ ions in the solution is equal to the sum of their individual concentrations.
Concentration of H+ ions:
[H+] = [HCl] + [HBr]
[H+] = 0.035 M + 0.080 M
[H+] = 0.115 M
The pH of a solution is given by the formula:
pH = -log[H+]
Calculating the pH:
pH = -log(0.115)
pH ≈ 0.94
Next, we can determine the pOH of the solution. The pOH is defined as the negative logarithm of OH- ion concentration.
The concentration of OH- ions can be calculated using the equation:
[OH-] = Kw / [H+]
Where Kw is the ion product of water and is equal to 1.0 x 10^-14 at room temperature.
Calculating the concentration of OH- ions:
[OH-] = Kw / [H+]
[OH-] = (1.0 x 10^-14) / (0.115)
[OH-] ≈ 8.7 x 10^-14 M
The pOH of a solution is given by the formula:
pOH = -log[OH-]
Calculating the pOH:
pOH = -log(8.7 x 10^-14)
pOH ≈ 13.06
Finally, we can find the pOH using the relationship between pH and pOH:
pH + pOH = 14
Calculating the pOH using the relationship:
pOH = 14 - pH
pOH ≈ 14 - 0.94
pOH ≈ 13.06
Therefore, the pH of the solution is approximately 0.94, and the pOH is approximately 13.06.
HCl is a strong acid. So is HBr. They ionize 100%. (H^+) from HCl = 0.035 moles/L. That from HBr is 0.080 moles/L. Add together to find total moles/L, then pH = -log(H^+). To find pOH, remember
pH + pOH = pKw = 14.