Question

10cm^3 of solution B is added to a beaker.
Distilled water is added to the beaker until the final volume in the beaker is 100cm^3
The pH of the solution is measured before and after distilled water is added.
table 4 sows the results(im jus going to write wats IN the table):

| Volume of solution in beaker: | Ph of solution: |
| 10cm^3 | 3 |
| 1000cm^3 | X |
Calculate the value of X

can someone help me on this question plz, i feel like its easy, but i just forgot how to do it lol

Answer 1

Answer:

4.0.

Explanation:

• From the given data that pH of 10 cm³ is 3.0, we can calculate its H⁺ concentration:

∵ pH = - log[H⁺] = 3.

∴ log [H⁺] = - 3.

∴ [H⁺] = 0.001 M.

• Now, we can calculate [H⁺] after the dilution process:

no. of millimoles before the dilution = no. of millimoles after the dilution

(MV) before dilution = (MV) after dilution

∴ M after dilution = [H⁺] = (MV) before dilution/ V after dilution = (10.0 cm³)(0.001 M)/(100.0 cm³) = 0.0001 M.

∴ pH of 100.0 cm³ solution = - log[H⁺] = - log(0.0001) = 4.0.