Answer:
See explanation below
Explanation:
At the beggining of the exercise, you already giving the solution for the 1st two problems, so I just going to answer the remaining question which are:
<em>"Determine the [ H + ] , [ OH − ] , and pOH of a solution with a pH of 7.20 at 25 °C. [ H + ] = M [ OH − ] = M pOH = 6.8 Determine the [ H + ] , [ OH − ] , and pH of a solution with a pOH of 6.43 at 25 °C."</em>
Now, to get every data in the problem, we need to write the expressions for every part, which are the following:
pH = -log[H⁺] (1)
pH = 14 - pOH (2)
[H⁺] = 10^(-pH) (3)
[H⁺] = Kw / [OH⁻] (4)
pOH = -log[OH⁻] (5)
pOH = 14 - pH (6)
[OH⁻] = 10^(-pOH) (7)
[OH⁻] = Kw / [H⁺] (8)
Now, according to the given data, we will use any of these eight expressions.
a) "<u><em>Determine the [ H + ] , [ OH − ] , and pOH of a solution with a pH of 7.20"</em></u>
We have pH, so in this case, we will use expression (3), (6) and (8):
[H⁺] = 10^(-7.20)
[H⁺] = 6.31x10⁻⁹ M
pOH = 14 - 7.2
pOH = 6.8
[OH⁻] = 1x10⁻¹⁴ / 6.31x10⁻⁹
[OH⁻] = 1.58x10⁻⁴ M
b) <u>"</u><u><em>The [ H + ] , [ OH − ] , and pH of a solution with a pOH of 6.43 "</em></u>
In this case, we can use expressions (2), (4) and (7). Using those expressions we have:
pH = 14 - 6.43
pH = 7.57
[H⁺] = 10^(-7.57)
[H⁺] = 2.69x10⁻⁸ M
[OH⁻] = 1x10⁻¹⁴ / 2.69x10⁻⁸
[OH⁻] = 3.72x10⁻⁷ M
, then solving for x, it is obtain that the fraction for 35 amu isotope is 0.775 (77.5%), and for 37 uma isotope is 0.225 (22.5%).