If you prepared a solution by adding 0.05mL (1drop) of 0.5M HCL to 5 x 10^7 mL of "pure" water. Calc the resultant PH. The answer isn't ph= 9.3. you have to add all the hydronium ions from all sources... but how ?
I would do this but check my thinking.
(HCl) added = 0.5M x (0.05/5E7)= 5E-10M
..............H2O ==> H^+ + OH^-
initial...............0......0
change.................x......x
add 5E-10M...........5E-10....x
final..............x+5E-10.....x
Kw = (H^+)(OH^-) = 1E-14
(x+5E-10)(x)=1E-14
Multiply out to obtain a quadratic that looks like this.
x^2 + 5E-10x - 1E-14 = 0
Solve the quadratic and I get something like 9.975E-8 = x. (I know that's more significant figures than we can legally use; however, if we round everything down to one place (from the 0.5M and the 0.05 mL) it would be 7 anyway and there would be no need to do any of this calculation.
Then x+5E-10 = ?? and pH from there. Close to 7 but less than 7 as it should be. It looks logical anyway.
To calculate the resultant pH of the solution after adding 0.05 mL of 0.5 M HCl to 5 x 10^7 mL of pure water, you need to consider the contribution of hydronium ions from both sources.
1. Calculate the amount of hydronium ions added from the HCl:
Since the volume of HCl added is 0.05 mL and the concentration is 0.5 M, you can calculate the moles of HCl added:
Moles of HCl = Volume (in L) x Concentration (in mol/L)
= 0.05 mL x (1 L/1000 mL) x 0.5 mol/L
= 0.000025 mol
HCl is a strong acid, so it dissociates completely, and each mole of HCl produces one mole of hydronium ions (H3O+). Therefore, the amount of hydronium ions added from HCl is also 0.000025 mol.
2. Calculate the amount of hydronium ions from the water:
The amount of hydronium ions in pure water can be determined using the equation for the dissociation of water:
H2O ⇌ H+ + OH-
In pure water, the concentration of hydronium ions (H+) is equal to the concentration of hydroxide ions (OH-), which can be calculated using the water dissociation constant, Kw:
Kw = [H+][OH-] = 1 x 10^-14
Since [H+] = [OH-], you can solve for [H+]:
[H+]^2 = 1 x 10^-14
[H+] = sqrt(1 x 10^-14)
[H+] ≈ 1 x 10^-7 mol/L
The volume of water is given as 5 x 10^7 mL, which is equivalent to 5 x 10^4 L. Therefore, the amount of hydronium ions from the water would be:
Hydronium ions from water = Concentration x Volume
= (1 x 10^-7 mol/L) x (5 x 10^4 L)
= 5 x 10^-3 mol
3. Total amount of hydronium ions:
To find the total amount of hydronium ions, you need to sum the values calculated from HCl and the water:
Total hydronium ions = Hydronium ions from HCl + Hydronium ions from water
= 0.000025 mol + 5 x 10^-3 mol
= 5.000025 x 10^-3 mol
4. pH calculation:
The pH is defined as the negative logarithm (base 10) of the hydronium ion concentration:
pH = -log[H+]
Therefore, to calculate the pH, you can substitute the value of [H+] into the equation:
pH = -log(5.000025 x 10^-3)
= -(-2.301)
= 2.301
Therefore, the resultant pH of the solution after adding 0.05 mL of 0.5 M HCl to 5 x 10^7 mL of pure water is approximately 2.301.
To calculate the resultant pH after adding the hydrochloric acid (HCl) to pure water, you need to consider the concentration of the hydronium ions (H3O+) in the resulting solution. Here's how you can calculate it:
1. Start with the given volume of water: 5 x 10^7 mL.
2. Convert it to liters by dividing by 1000: 5 x 10^7 mL ÷ 1000 = 5 x 10^4 L.
3. Calculate the number of moles of HCl added by using the molarity (0.5 M) and the volume (0.05 mL) in liters: 0.5 M x (0.05 mL ÷ 1000) L = 2.5 x 10^-6 moles of HCl.
4. Since hydrochloric acid is a strong acid and fully dissociates in water, the number of hydronium ions (H3O+) formed is equal to the number of moles of HCl added. Therefore, there are also 2.5 x 10^-6 moles of H3O+ ions.
5. Add the number of H3O+ ions from HCl to the initial concentration of H3O+ ions in pure water.
- In pure water, the concentration of H3O+ ions can be calculated using the formula pH = -log[H3O+], where [H3O+] is the concentration in moles per liter. The pH of pure water is 7 (neutral), which means the concentration of H3O+ ions is 1 x 10^-7 M.
- Adding the HCl, we have a resultant concentration of H3O+ ions: (1 x 10^-7 M) + (2.5 x 10^-6 moles ÷ 5 x 10^4 L) = 1 x 10^-7 M + 5 x 10^-5 M = 5 x 10^-5 M (rounded to two significant figures).
6. Calculate the pH of the resulting solution using the formula pH = -log[H3O+].
- pH = -log(5 x 10^-5) ≈ 4.3.
Therefore, the resultant pH after adding 0.05 mL of 0.5 M HCl to 5 x 10^7 mL of pure water is approximately pH 4.3.