Respuesta :
Answer:
Explanation:
Given parameters;
pH = 8.74
pH = 11.38
pH = 2.81
Unknown:
concentration of hydrogen ion and hydroxyl ion for each solution = ?
Solution
The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.
It is graduated from 1 to 14
pH = -log[H₃O⁺]
pOH = -log[OH⁻]
pH + pOH = 14
Now let us solve;
pH = 8.74
since pH = -log[H₃O⁺]
8.74 = -log[H₃O⁺]
[H₃O⁺] = 10⁻[tex]^{8.74}[/tex]
[H₃O⁺] = 1.82 x 10⁻⁹mol dm³
pH + pOH = 14
pOH = 14 - 8.74
pOH = 5.26
pOH = -log[OH⁻]
5.26 = -log[OH⁻]
[OH⁻] = 10[tex]^{-5.26}[/tex]
[OH⁻] = 5.5 x 10⁻⁶mol dm³
2. pH = 11.38
since pH = -log[H₃O⁺]
11.38 = -log[H₃O⁺]
[H₃O⁺] = 10⁻[tex]^{11.38}[/tex]
[H₃O⁺] = 4.17 x 10⁻¹² mol dm³
pH + pOH = 14
pOH = 14 - 11.38
pOH = 2.62
pOH = -log[OH⁻]
2.62 = -log[OH⁻]
[OH⁻] = 10[tex]^{-2.62}[/tex]
[OH⁻] =2.4 x 10⁻³mol dm³
3. pH = 2.81
since pH = -log[H₃O⁺]
2.81 = -log[H₃O⁺]
[H₃O⁺] = 10⁻[tex]^{2.81}[/tex]
[H₃O⁺] = 1.55 x 10⁻³ mol dm³
pH + pOH = 14
pOH = 14 - 2.81
pOH = 11.19
pOH = -log[OH⁻]
11.19 = -log[OH⁻]
[OH⁻] = 10[tex]^{-11.19}[/tex]
[OH⁻] =6.46 x 10⁻¹²mol dm³
1. For pH=8.74;
[H₃O⁺] = 1.82 x 10⁻⁹mol dm³
[OH⁻] = 5.5 x 10⁻⁶mol dm³
2. For pH=11.38;
[H₃O⁺] = 4.17 x 10⁻¹² mol dm³
[OH⁻] =2.4 x 10⁻³mol dm³
3. For pH=2.81;
[H₃O⁺] = 1.55 x 10⁻³ mol dm³
[OH⁻] =6.46 x 10⁻¹²mol dm³
It is given that the values of pH are as follows:
pH = 8.74
pH = 11.38
pH = 2.81
What is pH?
pH is a measure of the relative amount of free hydrogen and hydroxyl ions in the water.
pH formula:
pH = -log[[tex]H^{+}[/tex]] or pH = -log[[tex]OH^{-}[/tex]]
We know that, pH + pOH = 14
So, we need to calculate concentration of [H₃O⁺] and [OH⁻] for each of the following pH.
1. For pH=8.74
since pH = -log[H₃O⁺]
8.74 = -log[H₃O⁺]
[H₃O⁺] = [tex]10^{-8.74}[/tex]
[H₃O⁺] = 1.82 x 10⁻⁹mol dm³
pH + pOH = 14
pOH = 14 - 8.74
pOH = 5.26
pOH = -log[OH⁻]
5.26 = -log[OH⁻]
[OH⁻] = [tex]10^{-5.26}[/tex]
[OH⁻] = 5.5 x 10⁻⁶mol dm³
2. For pH=11.38
since pH = -log[H₃O⁺]
11.38 = -log[H₃O⁺]
[H₃O⁺] = [tex]10^{-11.38}[/tex]
[H₃O⁺] = 4.17 x 10⁻¹² mol dm³
pH + pOH = 14
pOH = 14 - 11.38
pOH = 2.62
pOH = -log[OH⁻]
2.62 = -log[OH⁻]
[OH⁻] = [tex]10^{-2.62}[/tex]
[OH⁻] =2.4 x 10⁻³mol dm³
3. pH=2.81
since pH = -log[H₃O⁺]
2.81 = -log[H₃O⁺]
[H₃O⁺] = [tex]10^{-2.81}[/tex]
[H₃O⁺] = 1.55 x 10⁻³ mol dm³
pH + pOH = 14
pOH = 14 - 2.81
pOH = 11.19
pOH = -log[OH⁻]
11.19 = -log[OH⁻]
[OH⁻] = [tex]10^{-11.19}[/tex]
[OH⁻] =6.46 x 10⁻¹²mol dm³
Learn more:
https://brainly.com/question/3775914
