Calculate the [OH-] of each aqueous solution with the following [H3O+].
coffee, 1.1x10^-5 M
Express your answer using two significant figures.
To find the [OH-], we can use the formula:
[H3O+][OH-] = 1.0 x 10^-14
Given that [H3O+] = 1.1 x 10^-5 M, we can rearrange the equation to solve for [OH-]:
[OH-] = (1.0 x 10^-14) / [H3O+]
= (1.0 x 10^-14) / (1.1 x 10^-5)
= 9.09 x 10^-10 M
Therefore, the [OH-] of the coffee solution is 9.09 x 10^-10 M.
To calculate the concentration of hydroxide ions ([OH-]) in an aqueous solution, we use the equation for the autoionization of water:
H2O ⇌ H+ + OH-
In neutral water at 25°C, the concentrations of hydronium ions ([H3O+]) and hydroxide ions ([OH-]) are equal: [H3O+] = [OH-] = 1.0 x 10^-7 M. However, in coffee, we are given a different concentration of hydronium ions ([H3O+]) as 1.1 x 10^-5 M.
To find the concentration of hydroxide ions ([OH-]), we can use the equation for the ion product of water (Kw):
Kw = [H3O+][OH-]
At 25°C, Kw is equal to 1.0 x 10^-14 M^2.
Rearranging the equation to solve for [OH-], we have:
[OH-] = Kw / [H3O+]
Substituting the values, we get:
[OH-] = (1.0 x 10^-14 M^2) / (1.1 x 10^-5 M)
[OH-] = 9.1 x 10^-10 M
Therefore, the concentration of hydroxide ions ([OH-]) in the coffee solution is approximately 9.1 x 10^-10 M (rounded to two significant figures).
To calculate the [OH-] of each aqueous solution, we need to use the relation between [H3O+] (hydronium ion concentration) and [OH-] (hydroxide ion concentration) in water. In pure water at 25°C, the concentration of [H3O+] is equal to the concentration of [OH-], which is 1.0 x 10^-7 M.
However, in this case, we are given the concentration of [H3O+] in coffee, which is an aqueous solution. We need to use the concept of pH to find the [OH-] concentration. The pH is defined as the negative logarithm (base 10) of [H3O+]. Mathematically, it is expressed as:
pH = -log[H3O+]
To find [OH-] from [H3O+], we can use the relationship:
[H3O+][OH-] = 1.0 x 10^-14 M^2
Rearranging the equation, we get:
[OH-] = 1.0 x 10^-14 M^2 / [H3O+]
Now, let's calculate the [OH-] for the given solution:
[H3O+] of coffee = 1.1 x 10^-5 M
Substituting this value into the equation, we have:
[OH-] = 1.0 x 10^-14 M^2 / (1.1 x 10^-5 M)
Simplifying the expression:
[OH-] = 9.09 x 10^-10 M
Therefore, the [OH-] concentration of the coffee solution is approximately 9.09 x 10^-10 M (rounded to two significant figures).