How do you calculate the pH, pOH, [H3O] and [OH] for a 0.067 M solution of sodium iodate
Sodium iodate is the salt of a strong base (NaOH) and a weak acid (HIO3). So this salt hydrolyzes as in the last problem I posted. The only difference is that this is a STRONG weak acid with Ka a relatively large value of about 0.18 or so. You need to confirm that. That means that you will not be able to neglect the y term; i.e.,
IO3^- + HOH ==> HIO3 + OH^-
Kb = Kw/Ka = (y)(y)/(0.067-y).
You must solve the quadratic because the concn of HIO3 and its Ka are so close together.
To calculate the pH, pOH, [H3O+], and [OH-] for a 0.067 M solution of sodium iodate, we need to consider the dissociation of the compound in water.
The chemical equation for the dissociation of sodium iodate (NaIO3) in water is:
NaIO3 (s) -> Na+ (aq) + IO3- (aq)
Since NaIO3 is a strong electrolyte, it completely dissociates in water, meaning we can assume that all of the sodium iodate ions separate into their respective ions when dissolved.
Step 1: Determine the concentrations of [H3O+] and [OH-].
In this case, because sodium iodate does not contain any hydrogen or hydroxide ions, we need to consider the autoionization of water to determine the concentration of [H3O+] and [OH-].
The autoionization of water in equilibrium is given by the equation:
H2O (l) ⇌ H3O+ (aq) + OH- (aq)
At 25 degrees Celsius, the concentration of [H3O+] and [OH-] in pure water is equal to 1.0 x 10^-7 M.
Step 2: Calculate the concentration of [H3O+] or [OH-] based on the chemical equation.
Since the sodium iodate dissociates fully into its respective ions, we can conclude that the concentration of [H3O+] and [OH-] will remain the same as they are in pure water.
Therefore, [H3O+] = 1.0 x 10^-7 M and [OH-] = 1.0 x 10^-7 M.
Step 3: Calculate the pH and pOH.
The pH is calculated by taking the negative logarithm (base 10) of the [H3O+]:
pH = -log [H3O+]
= -log (1.0 x 10^-7)
= 7
The pOH is calculated by taking the negative logarithm (base 10) of the [OH-]:
pOH = -log [OH-]
= -log (1.0 x 10^-7)
= 7
Step 4: Calculate the [H3O+] and [OH-].
[H3O+] can be calculated by taking the antilog of the negative pH value:
[H3O+] = 10^-pH
= 10^-7
= 1.0 x 10^-7 M
[OH-] can be calculated by taking the antilog of the negative pOH value:
[OH-] = 10^-pOH
= 10^-7
= 1.0 x 10^-7 M
So, for a 0.067 M solution of sodium iodate, the pH is 7, the pOH is 7, [H3O+] is 1.0 x 10^-7 M, and [OH-] is 1.0 x 10^-7 M.
To calculate the pH, pOH, [H3O+], and [OH-] for a given solution, you need to consider the dissociation of water and the relevant stoichiometry of the chemical equation.
1. Sodium iodate (NaIO3) is a strong electrolyte that completely dissociates in water. Thus, it will separate into its constituent ions:
NaIO3(s) → Na+(aq) + IO3-(aq)
2. Next, consider the dissociation of water, which results in the formation of hydronium ions ([H3O+]) and hydroxide ions ([OH-]):
H2O(l) ⇌ H3O+(aq) + OH-(aq)
3. For every one mole of NaIO3 that dissociates, it produces one mole each of Na+ and IO3-. However, since NaIO3 is a salt that does not affect the concentration of hydronium and hydroxide ions in solution, we can ignore it for pH and pOH calculations.
4. In a neutral solution, the concentration of [H3O+] and [OH-] would be equal: [H3O+] = [OH-] = 1.0 x 10^-7 M. However, in an acidic solution, [H3O+] is greater than [OH-], while in a basic solution, [OH-] is greater than [H3O+].
5. To determine the concentration of [H3O+] and [OH-] for a solution of sodium iodate, we need to know if the solution is neutral, acidic, or basic. This information can be deduced by calculating the pH and pOH.
6. The pH is calculated using the negative logarithm base 10 of [H3O+]: pH = -log[H3O+]. Similarly, the pOH is calculated using the negative logarithm base 10 of [OH-]: pOH = -log[OH-].
7. Since [H3O+] and [OH-] in a neutral solution are equal, the pH and pOH both equal 7.
In summary, the concentration of [H3O+] and [OH-] for a 0.067 M solution of sodium iodate can be determined by calculating the pH and pOH, which depend on the acidity or basicity of the solution.