Phosphoric acid is a triprotic acid:
H3PO4(aq) +H2O(l) ↔ H3O+(aq) + H2PO4-(aq) Ka1 = 7.5 x 10-3
H2PO4-(aq) +H2O(l) ↔ H3O+(aq) + HPO42-(aq) Ka2 = 6.2 x 10-8
HPO42-(aq) +H2O(l) ↔ H3O+(aq) + PO43-(aq) Ka3 = 4.8 x 10-13
Determine whether sodium monohydrogen phosphate (Na2HPO4) is neutral, basic, or acidic.
First, what is its Ka when it acts as an acid?
Second, what is its Kb when it acts as a base?
Na2HPO4, the salt, in aqueous solution is basic in that the pH > 7.0. WHY?
HPO4^2- + HOH --> H2PO4^- + OH^- and the OH^- provides the basicisity.
As an acid, HPO4^- + H2O ==> PO4^3- + H3O^+ and from the problem ka3 = 4.8E-13.
As a base the reaction is as above HPO4^2- + HOH --> H2PO4^- + OH^-
KaKb = Kw. You know Ka and Kw, solve for Kb. Kw = 1E-14 if you've forgotten.
Well, let's see how clown bot can help you out with this question!
To determine whether sodium monohydrogen phosphate (Na2HPO4) is neutral, basic, or acidic, we need to look at its Ka and Kb values.
Now, Na2HPO4 can act as an acid by donating a proton, and as a base by accepting a proton. Since we know the Ka values for the acid forms of phosphoric acid, we can use that information to find the Kb values for the corresponding base forms.
The Ka for the acid forms are as follows:
Ka1 = 7.5 x 10^-3
Ka2 = 6.2 x 10^-8
Ka3 = 4.8 x 10^-13
To find the Kb values, we can use the equation Ka x Kb = Kw (the water ionization constant). Since water is neutral, Kw = 1 x 10^-14.
Using this equation, we can find the Kb values for the base forms of phosphoric acid:
Kb1 = Kw / Ka1
Kb2 = Kw / Ka2
Kb3 = Kw / Ka3
Now, let's plug in the values and calculate!
Kb1 = (1 x 10^-14) / (7.5 x 10^-3)
Kb2 = (1 x 10^-14) / (6.2 x 10^-8)
Kb3 = (1 x 10^-14) / (4.8 x 10^-13)
And there you go! You have the Kb values for Na2HPO4 when it acts as a base. Now you can determine whether it's neutral, basic, or acidic.
Remember, though, that clown bot is here to bring some laughter to your day! So take these calculations with a grain of salt... or should I say, a pinch of circus powder? 😄
To determine whether sodium monohydrogen phosphate (Na2HPO4) is neutral, basic, or acidic, we need to consider its acid-base properties.
1. Ka when Na2HPO4 acts as an acid:
Na2HPO4 can act as an acid by donating a proton (H+). In this case, it can accept a hydroxide ion (OH-) to form the conjugate base HPO42- and water. The reaction is as follows:
Na2HPO4(aq) + H2O(l) ↔ HPO42-(aq) + OH-(aq)
Since we know Ka1 for H2PO4-, we can use the equilibrium expression to find Ka for HPO42-, which acts as an acid in this reaction:
Ka = [H3O+][HPO42-] / [Na2HPO4]
Substituting the known values from the given information:
Ka1 = 7.5 x 10^-3 (given)
[H3O+] = [H2PO4-] (since the concentration of H3O+ depends on the concentration of H2PO4-)
[HPO42-] = x (unknown)
[Na2HPO4] = concentration of Na2HPO4 (unknown)
As we don't know the concentrations of HPO42- and Na2HPO4, we cannot calculate the exact value for Ka when Na2HPO4 acts as an acid.
2. Kb when Na2HPO4 acts as a base:
In this case, Na2HPO4 can accept a proton (H+) from water and act as a base. The reaction is as follows:
Na2HPO4(aq) + H2O(l) ↔ HPO42-(aq) + H3O+(aq)
Since we know Ka2 for HPO42-, we can use the equilibrium expression to find Kb for Na2HPO4:
Kb = [H3O+][Na2HPO4] / [HPO42-]
Substituting the known values from the given information:
Ka2 = 6.2 x 10^-8 (given)
[H3O+] = [HPO42-] (since the concentration of H3O+ depends on the concentration of HPO42-)
[Na2HPO4] = x (unknown)
As we don't know the concentration of Na2HPO4, we cannot calculate the exact value for Kb when Na2HPO4 acts as a base.
In conclusion, we cannot determine the exact values for Ka or Kb because we don't have the necessary concentrations of Na2HPO4 and its conjugate base. However, based on the equilibrium constants and the given information, we can conclude that Na2HPO4 is a basic compound.
To determine whether sodium monohydrogen phosphate (Na2HPO4) is neutral, basic, or acidic, we need to look at its dissociation in water.
Na2HPO4 dissociates to produce Na+ ions and an HPO42- ion. The HPO42- ion can act as a base by accepting a proton (H+) or as an acid by donating a proton (H+).
To find the Ka when Na2HPO4 acts as an acid, we need to look at the dissociation reactions involving the HPO42- ion:
HPO42-(aq) + H2O(l) ↔ H3O+(aq) + PO43-(aq) (Equation 1)
The Ka expression for Equation 1 can be written as:
Ka = [H3O+][PO43-] / [HPO42-]
Given that Ka3 = 4.8 x 10^-13 for this reaction, we can substitute the concentrations of the ions at equilibrium into the Ka expression to find the value of Ka.
To find the Kb when Na2HPO4 acts as a base, we need to consider the reaction where HPO42- accepts a proton:
HPO42-(aq) + H2O(l) ↔ H3O+(aq) + PO43-(aq) (Equation 2)
The Kb expression for Equation 2 can be written as:
Kb = [H3O+][PO43-] / [HPO42-]
Since Kb = Kw / Ka, we can use the value of Ka3 and the autoionization constant of water Kw to find the Kb value.
Kw = [H3O+][OH-] = 1.0 x 10^-14 at 25°C
Kw = Ka1 * Ka2 * Ka3
We can rearrange the equation to solve for Kb:
Kb = Kw / Ka = (1.0 x 10^-14) / Ka3
Now we have the expressions for both Ka and Kb when Na2HPO4 acts as an acid or a base.