Question

Be sure to answer all parts. The molecular scene at right depicts the relative concentrations of H3O+ (purple) and OH− (green) in an aqueous solution at 25 ° C. (Counter ions and solvent molecules are omitted for clarity.) (a) Calculate the pH. $sd_vs17_green.png$ (b) How many H3O+ ions would you have to draw for every OH− ion to depict a solution of pH 4.90? Enter your answer in scientific notation. $sd_vs17_red.png$ × 10 $sd_vs17_red.png$

Answer 1

Solution:

Part A:

Number of H₃O⁺ ions, n₁ = 2

Number of OH^- ions, n₂ = 20

Ratio of their concentrations,

[H3O+] [OH-] n1 na

[H30+] 2 [OH-] 20

T0 = -HO) [+o®H]

We know that

K_w = [H₃O⁺] × [OH^-​​​​​​]

10^-14 =  [H₃O⁺] × [OH^-​​​​​​]

10^-14 = [H₃O⁺] × ([H₃O⁺​​​​​​]/0.1)

([H₃O⁺])² = 10¹⁵

[H₂O⁺] = 0.00000003162

pH = - log [H₃O⁺]

pH = - log(0.00000003162)

pH = 7.50

Part B:

pH of the solution = 4.90

We know that

pH + pOH = 14

4.90 + pOH = 14

pOH = 9.1

or - log[OH^-​​​​​​] = 9.1

[OH^-] = 10^-9.1

[OH^-] = 794.328234724 x 10^-12

Also,

K_w = [H₃O⁺] × [OH^-​​​​​​]

10^-14 =  [H₃O⁺] × [OH^-​​​​​​]

10^-14 = [H₃O⁺] × (7.94328 x 10^-10)

[H₃O⁺] = 0.00001258925

Therefore,

[H3O+] OH- 1 0.00001258925 794.328234724 * 10-12

[H30*] 2 = 15848.9267 ОН-

[H30*] он-1 = 1,58489 х 10

1.58489 x 10⁴ ions would have to be draw forevery OH^-for the given solution.