Will the 90-100 mL of distilled water added to the acid solution (10.00 mL) significantly change the moles of H3O+ in the 1.00 M HCl solution? Why or why not?
It won't change moles H3O^+ significantly because H2O has few H3O^+ to add; however, it will change the mols/L. The new M will be 1 x 10/100 = ?.
To determine whether the addition of 90-100 mL of distilled water will significantly change the moles of H3O+ in the 1.00 M HCl solution, we need to consider dilution.
Dilution is a process that involves adding solvent to a solution to decrease its concentration. In this case, we are adding distilled water to the 1.00 M HCl solution.
To calculate the new concentration, we can use the dilution formula:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution
V1 = initial volume of the solution
C2 = final concentration of the solution
V2 = final volume of the solution
Given that the initial concentration (C1) is 1.00 M and the initial volume (V1) is 10.00 mL, we can rearrange the formula to solve for the final concentration (C2):
C2 = (C1 * V1) / V2
Now, let's calculate the final concentration when 90-100 mL of distilled water is added. Assuming the final volume (V2) is 90-100 mL + 10.00 mL = 100-110 mL, we can calculate the final concentration (C2):
C2 = (1.00 M * 10.00 mL) / (100-110 mL)
C2 = 0.10-0.091 M
From the calculations, we can see that the final concentration of the HCl solution will be significantly lower after the dilution.
To determine whether the addition of the distilled water significantly changes the moles of H3O+ in the solution, we need to compare the moles of H3O+ initially present in the 10.00 mL of 1.00 M HCl to the moles of H3O+ in the final diluted solution.
The number of moles can be calculated using the equation:
moles = concentration (M) * volume (L)
Initially, in the 10.00 mL of 1.00 M HCl, we can calculate the moles of H3O+:
moles_initial = 1.00 M * (10.00 mL / 1000 mL/L)
moles_initial = 0.01 moles
In the final diluted solution, we can calculate the moles of H3O+ using the final concentration (C2) and the final volume (V2):
moles_final = C2 * (V2 / 1000 mL/L)
moles_final = (0.10-0.091 M) * (100-110 mL / 1000 mL/L)
moles_final = 0.0091-0.0101 moles
Comparing the moles_initial (0.01 moles) to the moles_final (0.0091-0.0101 moles), we can observe that the change in moles is minimal.
Therefore, adding 90-100 mL of distilled water to the 10.00 mL of 1.00 M HCl solution does not significantly change the moles of H3O+.
To determine whether adding 90-100 mL of distilled water to a 10.00 mL solution of 1.00 M HCl will significantly change the moles of H3O+ present, we need to apply some calculations. Let's break it down step by step:
1. Calculate the moles of H3O+ in the original 10.00 mL, 1.00 M HCl solution:
- To find the moles of a substance, use the formula: Moles = Concentration × Volume.
- Moles = 1.00 M × 0.0100 L = 0.0100 moles
2. Determine the new total volume after adding 90-100 mL of distilled water:
- Original volume + Distilled water volume = Total volume
- 10.00 mL + 90-100 mL = 100.00 mL
3. Convert the total volume to liters:
- 100.00 mL ÷ 1000 mL/L = 0.1000 L
4. Calculate the new concentration of HCl after dilution:
- Concentration = Moles ÷ Volume
- New concentration = 0.0100 moles ÷ 0.1000 L = 0.100 M
Now, let's analyze the results:
By diluting the 10.00 mL, 1.00 M HCl solution by adding 90-100 mL of distilled water, the moles of H3O+ stay the same at 0.0100 moles. However, the total volume of the solution increases to 100.00 mL or 0.1000 L. As a result, the new concentration is 0.100 M.
Therefore, adding the distilled water does not significantly change the moles of H3O+ present, but it does decrease the concentration of the HCl solution.