What is the pH of a 0.100 molar HCl solution?

Answer 1

The answer depends on several unspecified variables, most importantly the final molarity of the solution, which depends on the final volume.

You can calculate the value yourself using the formula:

pH = -log[H+]

where [H+] is the final concentration of H+ ions in solution. For HCl, [H+] is equal to molarity.

So, for example, if you add 50.0 ml of 1.0M HCl to 950 ml of deionized water, your final concentation is:

(50.0 ml/1000 ml) * (1.0M) = 0.05M

Therefore:

pH = -log[0.05] = 1.3

Answer 2

If your question was something like this: 100.mL of 0.200M HCl is titrated with 0.250M NaOH.

A) What is the pH of the solution after 50.0mL of base has been added?

B)What is the pH of the solution at the equivalence point?

A) moles HCl = 100 x 0.2 /1000 = 0.02

Moles NaOH = 50 x 0.250 / 1000 = 0.0125

moles HCl in excess = 0.02 - 0.0125 = 0.0075

Total volume = 150 mL = 0.150 L

[H+] = 0.0075 / 0.150 = 0.05 M

pH = - log 0.05 = 1.30

B) at equivalence point [OH-] = [H+] so pH = 7

Answer 3

.003 L HCL * 2.5 moles HCL/1L = .0075 moles HCL

.0075 moles HCL/ .1 L HCL = .075 M HCL = .075 M H3O+

pH = -log[H3O+]

pH = -log[.075] = 1.12

this of course is not adjusted for activities of the ions in solution, but it would be very close to this value.

Answer 4

The pH of a solution made by mixing 40.00 mL of 0.10 M HCl with 25.00 mL of 0.10 M KOH:

Here the molarity of base and acid are same but the volume of acid is more than the volume of base so the resultant solution is acidic.

The concentration of OH- in solution = (molarity of base * volume of base - molarity of acid * volume of acid) / total volume

= (40 mL*0.1M - 25 mL *0.1 M ) / (40 mL +25mL)/ 1000)

= 0.021428571 M

pH = -log(0.021428571) = 1.669

Answer 5

It should be some where around .10. But this is a conjecture as the correspondent has not had Chemistry for many years and the answers provided on the net are not very clear.

Answer 6

- log(0.100 M HCl)

= 1 pH
=====

Answer 7

1.70

Answer 8

.17