For each of the following solutions, calculate the initial pH and the final pH after adding 0.010 mol
of HCℓ.
a) 500.0 mL of pure water
i'm not sure how to do this question..
would the equation be
h20+h30 equilibrium arrows h30+h20
You are supposed to recognize that HCl is a strong acid (meaning it ionizes 100%) so pure water is pH of 7. Then you add 0.01 mol HCl which adds to 500 mL water. That 0.01 mol becomes 0.01/0.5 L = ?? and convert that to pH.
Well, it looks like there's some confusion there. The equation you provided is not balanced and doesn't quite capture what's happening in this scenario.
When you add hydrochloric acid (HCl) to water, it dissociates into hydrogen ions (H+) and chloride ions (Cl-). This can be represented by the balanced equation:
HCl + H2O → H3O+ + Cl-
Now, let's move on to the calculations. Since you're adding 0.010 mol of HCl to 500.0 mL of pure water, we need to convert the volume to liters for the calculations. 500.0 mL is equivalent to 0.5000 L.
To find the initial pH, we need to determine the concentration of H3O+ ions in the solution. Since HCl is completely ionized in water, the concentration of H3O+ ions will be equal to the concentration of HCl you added.
Initial concentration of H3O+ ions = 0.010 mol / 0.5000 L = 0.020 M
Using the equation pH = -log[H3O+], we can find the initial pH:
Initial pH = -log(0.020) ≈ 1.70
Now, let's calculate the final pH after adding the HCl. Since HCl is a strong acid, it completely dissociates, so the concentration of H3O+ ions will double in the solution.
Final concentration of H3O+ ions = 2 * 0.020 M = 0.040 M
Final pH = -log(0.040) ≈ 1.40
So, the initial pH is approximately 1.70, and the final pH is approximately 1.40.
Yes, the equation you wrote is the dissociation of water:
H2O + H3O+ ⇌ H3O+ + OH-
In pure water, the concentration of H3O+ (also known as H+) and OH- ions are equal and can be calculated using the autoionization constant (Kw) of water which is equal to 1.0 x 10^-14 at 25°C.
Since the solution is pure water, the initial concentration of H+ ions (H3O+) is equal to the initial concentration of OH- ions, which can be calculated as:
[H3O+] = [OH-] = √(Kw)
Given that Kw = 1.0 x 10^-14:
[H3O+] = [OH-] = √(1.0 x 10^-14) = 1.0 x 10^-7 M
Now, we can calculate the initial pH using the definition of pH:
pH = -log[H3O+]
pH = -log(1.0 x 10^-7)
pH = 7.0
When 0.010 moles of HCl is added, it completely dissociates into H+ and Cl- ions. Since the initial concentration of H+ ions was 1.0 x 10^-7 M and 0.010 moles of HCl has been added to the solution, the new concentration of H+ ions can be calculated as:
[H3O+]final = [H3O+]initial + [H+ from HCl]
[H3O+]final = 1.0 x 10^-7 M + 0.010 mol / 0.500 L
[H3O+]final = 1.0 x 10^-7 M + 2.0 x 10^-2 M
[H3O+]final = 2.0 x 10^-2 M
Finally, we can calculate the final pH using the new concentration of H+ ions:
pHfinal = -log[H3O+]final
pHfinal = -log(2.0 x 10^-2)
pHfinal ≈ 1.70
To calculate the initial and final pH after adding 0.010 mol of HCℓ to 500.0 mL of pure water, we need to take into account the dissociation of HCℓ in water and the autoprotolysis of water.
Firstly, the dissociation equation for HCℓ in water is:
HCℓ ⇌ H⁺ + Cl⁻
The concentration of H⁺ ions can be determined from the moles of HCℓ added and the volume of the solution:
C(H⁺) = (moles of HCℓ) / (volume of solution)
In this case, the moles of HCℓ added is 0.010 mol and the volume of the solution is 500.0 mL (which can be converted to 0.5000 L):
C(H⁺) = 0.010 mol / 0.5000 L = 0.020 M
Since HCℓ is a strong acid, it will dissociate completely, meaning that the concentration of H⁺ ions in solution will be equal to the concentration of HCℓ.
Now, let's calculate the initial pH using the formula:
pH = -log[H⁺]
pH = -log(0.020) = 1.70
This is the initial pH before any reaction takes place.
Since we added an acidic solution to pure water, the concentration of H⁺ ions will increase. The final pH can be calculated using the Henderson-Hasselbalch equation:
pH = -log([H⁺]) = -log(Kw / [OH⁻])
In pure water, the concentration of H⁺ ions and OH⁻ ions are equal and can be determined using the equation for the autoprotolysis of water:
2H₂O ⇌ H₃O⁺ + OH⁻
For pure water, [H⁺] = [OH⁻] = 1.00 x 10⁻⁷ M (at 25°C).
Now, let's calculate the final concentration of H⁺ ions after adding HCℓ:
[H⁺] = [OH⁻] + [C(H⁺)]
[H⁺] = 1.00 x 10⁻⁷ M + 0.020 M = 0.020001 M
pH = -log(0.020001) = 1.70
Therefore, the final pH after adding 0.010 mol of HCℓ to 500.0 mL of pure water remains the same as the initial pH and is equal to 1.70.