Chemistry
The hydronium ion concentration of lemon juice is 5 x 10^ –3.
A sample of lemon juice will have [OH–] value of ___i___ mol/L and pH value of ___ii___. The statement given above is completed by the information in row:
Row i ii
A 5 x 10^–12 4.2
B. 3 x 10^–12 3.1
C. 2 x 10^–12 2.3
D. 2 x 10^–11 1.2
E. 4 x 10^–12 2.7
a) A
b) B
c) C
d) D
e) E
The concentration of hydroxide ions ([OH-]) can be calculated using the formula Kw = [H3O+][OH-], where Kw is the ion product of water (1 x 10^-14 mol^2/L^2). Rearranging the equation, [OH-] = Kw / [H3O+].
Given that the hydronium ion concentration of lemon juice (H3O+) is 5 x 10^-3 mol/L, we can calculate the value of [OH-]:
[OH-] = (1 x 10^-14 mol^2/L^2) / (5 x 10^-3 mol/L)
[OH-] = 2 x 10^-12 mol/L
Therefore, the value of [OH-] is 2 x 10^-12 mol/L.
To calculate the pH value, we can use the formula pH = -log[H3O+]. Substituting the given value of hydronium ion concentration, we get:
pH = -log(5 x 10^-3)
pH = 2.3
Therefore, the pH value of the lemon juice sample is 2.3.
From the information provided, we can conclude that the completed statement is given by option C: C. 2 x 10^–12 (i) and 2.3 (ii).
are you sure?
Apologies for the mistake in my previous response. Let me recalculate and provide you with the correct answer.
Given that the hydronium ion concentration of lemon juice (H3O+) is 5 x 10^-3 mol/L, we can calculate the value of [OH-]:
[OH-] = (1 x 10^-14 mol^2/L^2) / (5 x 10^-3 mol/L)
[OH-] = 2 x 10^-12 mol/L
Therefore, the value of [OH-] is 2 x 10^-12 mol/L.
To calculate the pH value, we can use the formula pH = -log[H3O+]. Substituting the given value of hydronium ion concentration, we get:
pH = -log(5 x 10^-3)
pH ≈ 2.3
Therefore, the pH value of the lemon juice sample is approximately 2.3.
Upon reconsideration, the correct completed statement is given by option C: C. 2 x 10^–12 (i) and 2.3 (ii).
To find the [OH–] value and pH value of a sample of lemon juice, we need to use the relationship between [H3O+] (hydronium ion concentration) and [OH–] (hydroxide ion concentration) in water.
In pure water at 25°C, the concentrations of [H3O+] and [OH–] are equal, each being 1 x 10^–7 mol/L. This is known as the neutral pH value of 7, where [H3O+] = [OH–].
However, lemon juice is acidic, so it has a higher concentration of [H3O+] ions (hydronium ions) and a lower concentration of [OH–] ions (hydroxide ions).
Given that the hydronium ion concentration of lemon juice is 5 x 10^–3 mol/L, we can use this information to find the [OH–] value and pH value.
To find the [OH–] value, we can use the equation:
[H3O+] x [OH–] = 1 x 10^–14 (at 25°C)
Substituting the given [H3O+] value of 5 x 10^–3 mol/L:
(5 x 10^–3) x [OH–] = 1 x 10^–14
Solving for [OH–]:
[OH–] = (1 x 10^–14) / (5 x 10^–3)
[OH–] ≈ 2 x 10^–12 mol/L
So, the [OH–] value of the lemon juice sample is approximately 2 x 10^–12 mol/L.
Next, to find the pH value, we can use the equation:
pH = -log[H3O+]
Substituting the given [H3O+] value:
pH = -log(5 x 10^–3)
pH ≈ 2.3
So, the pH value of the lemon juice sample is approximately 2.3.
Comparing these calculated values to the options given, we can see that they match up with option C:
Row i: 2 x 10^–12 (approx.)
Row ii: 2.3 (approx.)