suppose that after measuring the pH of 1.05 M HCl solution, three drops (0.20 mL) of the solution remained behind on the pH electrode.
a) when this electrode is then dipped into 100 mL of pure distilled water, show by calculation that the pH of the water is not pH= 7
HCl = 1.05 x 0.2/100 = about 0.0021M
..........H2O ==> H^+ + OH^-
I...............0.0021...0
C.................+x.....+x
E.............0.0021-x....x
Kw = (H^+)(OH^-)
1E-14 = (0.0021+x)(x)
Solve for x and convert H^+ to pH.
I don't think the ionization of H2O even enters into it.
To calculate the pH of the water after the electrode has been dipped into it, we need to consider the dilution caused by mixing the drops of HCl solution with the water. Here's how you can calculate it:
1. Calculate the amount of HCl that remains in the drops:
- Since the initial concentration of HCl solution is 1.05 M and the volume of the drops is 0.20 mL, the amount of HCl in the drops is given by:
Amount of HCl = concentration × volume
= 1.05 M × 0.20 mL
= 0.21 mmol
2. Calculate the dilution factor caused by mixing the drops with the water:
- The final volume of the solution is 100 mL (water) + 0.20 mL (drops) = 100.20 mL.
- The dilution factor is the ratio of the final volume to the initial volume:
Dilution factor = final volume / initial volume
= 100.20 mL / 0.20 mL
= 501
3. Calculate the molarity of HCl in the diluted solution:
- Since the amount of HCl remains the same, we can use the dilution factor to calculate the new concentration:
Concentration = Amount of HCl / final volume
= 0.21 mmol / 100.20 mL
= 2.10 × 10^−3 M
4. Calculate the pH of the diluted HCl solution:
- The pH of a strong acid like HCl can be calculated using the formula:
pH = -log[H+]
where [H+] is the concentration of hydrogen ions.
- In this case, the concentration of H+ is the same as the concentration of HCl:
[H+] = 2.10 × 10^−3 M
- Hence, the pH of the diluted HCl solution is:
pH = -log(2.10 × 10^−3)
≈ 2.68
Therefore, after dipping the pH electrode that contained 3 drops of 1.05 M HCl solution into 100 mL of pure distilled water, the pH of the water is approximately 2.68 and not pH = 7 as expected for pure water.