Sodium Azide, (NaN3), the conjugate base of hydrazoic acid, is sometimes added to water to kill bacteria. Given a 0.010 M solution of sodium azide,
A) What is the [H+] concentration from the Azide base rxn?
B) Given a 0.01 M NH4+, what is the [H+]?
C) What is the hydrozoic acid (NH3) half cell potential given 1atm N2(g)?
Please show steps
A. One note on this part. NaN3 is NOT the conjugate base of HN3. N3^- is the conjugate base. (NaN3 is the salt of the conjugate base.)
...........N3^- + HOH ==> HN3 + OH-
I........0.01.............0......0
C.........-x..............x......x
E.......0.01-x............x......x
Kb for N3^- = (Kw/Ka for HN3) = (x)(x)/(0.01-x). Solve for x = (OH^-) and convert to H^+.
B. I assume you mean NH4^+ and not ammonium azide. This is done the same way as part A but the hydrolysis is different.
NH4^+ + H2O ==> NH3 + H3O^+
Ka for NH4^+ = (Kw/Kb for NH3) = etc.
C. Can't you find this value in tables?
coppers atomic mass is 63.5 and occurs in nature in the form of twi isotopes Cu-63 abs Cu 65
A) Well, I must admit, adding a little NaN3 to water is certainly one way to deal with bacteria. So, let's calculate the [H+] concentration from the Azide base reaction.
First, we need to write the chemical equation for the reaction between NaN3 and water:
NaN3 + H2O -> NH3 + NaOH
From the balanced equation, we can see that 1 mole of NaN3 produces 1 mole of NH3 and 1 mole of NaOH. Since sodium hydroxide (NaOH) is a strong base, it will dissociate fully to produce OH- ions. Therefore, we don't need to consider it in our calculations.
Now, let's calculate the [H+] concentration. Since we have a 0.010 M solution of sodium azide (NaN3), the concentration of NH3 formed will also be 0.010 M.
[H+] concentration from the Azide base reaction = [NH3] = 0.010 M
B) Moving on to the next question, given a 0.01 M NH4+, we need to calculate the [H+] concentration. Oh, the wonderful world of chemistry!
Now, you don't need to be a clown like me to solve this. You just need to remember that ammonium (NH4+) is actually an acid. It can donate a proton (H+) to water to form hydronium ions (H3O+).
So, the equation for the reaction between NH4+ and H2O is:
NH4+ + H2O -> H3O+ + NH3
From the balanced equation, we can see that 1 mole of NH4+ produces 1 mole of NH3 and 1 mole of H3O+.
Now, let's calculate the [H+] concentration. Since we have a 0.01 M NH4+ solution, the concentration of NH3 formed will also be 0.01 M.
[H+] concentration from NH4+ = [NH3] = 0.01 M
C) Ah, the half-cell potential of hydrazoic acid (NH3). That's a good one!
First, let's write the half-reaction equation:
NH3 -> 1/2 N2 + 3/2 H+
Now, we need to find the standard reduction potential for this reaction. Unfortunately, I don't have that information right now. But hey, don't be blue! You can find it in a table of standard reduction potentials.
Once you have the standard reduction potential, you can use the Nernst equation to calculate the half-cell potential under any given conditions. Just don't forget to take into account the concentration of the species involved and the temperature.
A) First, let's write out the balanced equation for the dissociation of sodium azide in water:
NaN3 + H2O -> Na+ + N3- + H2O
Since sodium azide dissociates into Na+ and N3- ions, and N3- acts as the conjugate base of hydrazoic acid (HN3), we can determine the concentration of [H+] using the reaction:
N3- + H2O ⇌ HN3 + OH-
From the balanced equation, we can see that for every N3- ion that reacts with water, one HN3 molecule is formed. Therefore, the concentration of [H+] is equal to the concentration of NaN3.
So, the [H+] concentration from the Azide base reaction is 0.010 M.
B) For the dissociation of ammonium (NH4+) in water, we have the reaction:
NH4+ + H2O ⇌ NH3 + H3O+
From the balanced equation, we can see that for every NH4+ ion that reacts with water, one H3O+ ion is formed. Therefore, the concentration of [H+] is equal to the concentration of NH4+.
So, the [H+] concentration from the 0.01 M NH4+ is also 0.01 M.
C) The half-cell potential for the reaction HN3 ⇌ NH3 + 1/2N2(g) can be determined using the Nernst equation:
E = E° - (RT/nF) * ln(Q)
Where:
E = cell potential
E° = standard cell potential
R = gas constant (8.314 J/(mol∙K))
T = temperature in Kelvin
n = number of electrons transferred in the reaction
F = Faraday's constant (96485 C/mol)
Q = reaction quotient
The E° value for the half-cell reaction can be looked up in a table. Let's assume that the E° value for this reaction is -0.62 V.
Let's also assume that the temperature is 298 K.
To calculate Q, we need the concentrations of NH3 and N2. We know that 1 atm N2(g) is provided, so the concentration of N2 is 1 atm. However, the concentration of NH3 is not given, so we cannot calculate Q or the cell potential accurately.
Therefore, we do not have enough information to determine the half-cell potential (H2N3/NH3) given 1 atm N2(g).
A) To find the [H+] concentration from the Azide base reaction, we need to write the balanced chemical equation for the reaction between sodium azide (NaN3) and water (H2O):
2 NaN3 + 2 H2O -> 2 NH3 + 3 H2O + N2
From the equation, we can see that two sodium azide molecules react with two water molecules to produce two ammonia (NH3) molecules, three water molecules, and one molecule of nitrogen gas (N2).
Since NaN3 is the conjugate base of hydrazoic acid, the azide ion (N3-) acts as a base and accepts a proton (H+) from water to form ammonia (NH3):
N3- + H2O -> NH3 + OH-
In this reaction, one azide ion reacts with one water molecule to produce one ammonia molecule and one hydroxide ion.
From the balanced equation, we can see that the molar ratio between azide ions (N3-) and hydroxide ions (OH-) is 1:1. This means that the [OH-] concentration is equal to the [N3-] concentration.
Given that the concentration of sodium azide (NaN3) is 0.010 M, the [N3-] concentration is also 0.010 M.
Therefore, [OH-] = [N3-] = 0.010 M.
Since water dissociates into equal concentrations of hydrogen ions (H+) and hydroxide ions (OH-), the [H+] concentration can be calculated using the Kw (ionization constant of water) equation: Kw = [H+][OH-].
Kw is a constant at a given temperature, and at 25°C, its value is 1 × 10^-14.
We know the [OH-] concentration is 0.010 M, so we can substitute these values into the Kw equation to solve for the [H+] concentration:
(1 × 10^-14) = [H+](0.010)
[H+] = (1 × 10^-14) / (0.010) = 1 × 10^-12 M.
Therefore, the [H+] concentration from the Azide base reaction is 1 × 10^-12 M.
B) To find the [H+] concentration from the 0.01 M NH4+ solution, we need to consider the reaction between ammonium (NH4+) and water (H2O). The balanced equation is:
NH4+ + H2O -> NH3 + H3O+
From the equation, we can see that an ammonium ion reacts with water to produce ammonia (NH3) and a hydronium ion (H3O+).
Given that the concentration of NH4+ is 0.01 M, the [NH4+] concentration is also 0.01 M.
Since the molar ratio of NH4+ to H3O+ is 1:1, the [H3O+] concentration is also 0.01 M.
Hence, the [H+] concentration is equal to the [H3O+] concentration and is also 0.01 M.
Therefore, the [H+] concentration from the 0.01 M NH4+ solution is 0.01 M.
C) To find the half-cell potential (E) for the hydrazoic acid (NH3) half-cell reaction, we need to find the standard reduction potential (E°).
The half-cell reaction is:
NH3 + 3H2O + 4e- -> NH4+ + 5OH-
The standard reduction potential (E°) is a measure of the tendency of a species to gain electrons. It can be found in tables of standard reduction potentials.
Given that the half-cell reaction is the reduction of NH4+ to NH3, we can find the E° value for this reaction.
Using the Nernst equation: E = E° - (0.0592/n) * log([NH3]/[NH4+]), where n is the number of electrons transferred in the half-cell reaction, we can calculate the half-cell potential (E).
At standard conditions (25°C, 1 atm, 1 M), the balanced equation is:
NH4+ + OH- -> NH3 + H2O + e-
The E° value for this reaction is given as a standard reduction potential. You can refer to a table of standard reduction potentials to find the specific E° value for this reaction.
Once you have the E° value, substitute it into the Nernst equation along with the concentrations of NH3 and NH4+, and solve for E.
Note that since the problem states "1 atm N2(g)", it implies that the reaction is occurring under gas conditions, so the use of the Nernst equation is appropriate.
Therefore, the steps to calculate the half-cell potential (E) for the hydrazoic acid (NH3) half-cell reaction involve referring to a table of standard reduction potentials to find the E° value and then using the Nernst equation to calculate E.