You start with some \"pure\" water at 25.0° C (a temperature at which the value

Question

You start with some "pure" water at 25.0° C (a temperature at which the value of Kw = 1.01 × 10-14) and add enough HCl so that the resulting solution is 5.00 × 104 M HCl. What is the value of the pH? ["Pure" water has no solutes dissolved in it and has been degassed. Before you add the HCl, there is only H2O and its dissociation products.] (2 pts) 2. a. You start with some "pure" water at 25.0° C (a temperature at which the value of Kw = 1.01 × 10-14) and add enough NaOH so that the resulting solution is 7.00 x 105 M NaOH. What is the value of the pH? (2 pts) b. 1.01 × 10-14) and add You start with some "pure" water at 25.0° C (a temperature at which the value of Kw enough HCl so that the resulting solution is 6.00 × 10-9 M HCl. What is the value of the pH7 (3 pts) c.

Explanation / Answer

a)   [H3O+] = 5.00 x 10^-4 M

pH = -log [H3O+]

pH = -log (5.00 x 10^-4)

pH = 3.30

b)

[OH-] = 7.00 x 10^-5 M

[H3O+] [OH-] = Kw

[H3O+] x 7.00 x 10^-5   = 1.01 x 10^-14

[H3O+] = 1.44 x 10^-10 M

pH = -log [H3O+]

pH = 9.84

c)

[H3O+] = [H3O+] from HCl + [H3O+] from water

              = 6.00 x 10^-9 + 1.00 x 10^-7

                = 1.06 x 10^-7 M

pH = 6.97

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.