An aqueous solution has a pOH = 6.5. What is the [H3O+] of this solution?
pH + pOH = pKw = 14.
You know pKw = 14 and you know pOH. calculate pH, then
pH = -log(H3O^+) Solve for H3O^+
To find the [H3O+] of the solution, we need to use the relationship between pOH and pH.
Recall that pOH + pH = 14.
Given pOH = 6.5, we can find pH by subtracting pOH from 14:
pH = 14 - pOH
= 14 - 6.5
= 7.5
The pH of the solution is 7.5.
Next, we can find [H3O+] using the relationship between pH and [H3O+].
Recall that pH is the negative logarithm (base 10) of the [H3O+] concentration:
pH = -log[H3O+]
Rearranging the equation, we have:
[H3O+] = 10^(-pH)
Substituting the pH value we found earlier:
[H3O+] = 10^(-7.5)
Now, we can calculate [H3O+] using a scientific calculator or by performing the calculation step by step:
[H3O+] = 0.000003162
Therefore, the [H3O+] of the solution is approximately 0.000003162.
To find the [H3O+] concentration of the solution, we can use the relationship between pOH and [OH-]. The formula for pOH is:
pOH = -log[OH-]
Given pOH = 6.5, to find the [OH-] concentration, we need to convert pOH into [OH-]. We use the following formula:
[OH-] = 10^(-pOH)
Substituting the given value:
[OH-] = 10^(-6.5)
Now, to find the [H3O+] concentration, we can apply the concept of the ion product of water (Kw). At 25°C, Kw = [H3O+][OH-] = 1.0 x 10^(-14).
Given that [OH-] = 10^(-6.5), we can rearrange the equation to solve for [H3O+]:
[H3O+] = Kw / [OH-]
[H3O+] = (1.0 x 10^(-14)) / (10^(-6.5))
Simplifying:
[H3O+] = (1.0 x 10^(-14)) * (10^(6.5))
Now, we can use the logarithmic property: log a^b = b*log a. Applying this property:
[H3O+] = (1.0 x 10^(-14)) * (10^(6.5))
[H3O+] = (1.0 x 10^(-14)) * (10^(6.5))
[H3O+] ≈ 3.16 x 10^(-9)
Therefore, the concentration of [H3O+] in the solution is approximately 3.16 x 10^(-9) M.