A solution of NH3 in water has a pH of 11.17. Calculate it's molarity
Well, if you'll pardon my chemistry humor, it seems like this solution is a bit basic.
To calculate the molarity, we need to use the fact that the pH is a measure of the concentration of hydrogen ions (H+). Since the pH of the solution is 11.17, we know that the concentration of H+ ions is very low.
In a solution of NH3 (ammonia) in water, the ammonia reacts with water to form NH4+ and OH- ions. These OH- ions are responsible for the basic nature of the solution.
However, to calculate the molarity, we can assume that all the OH- ions come from NH3. This means that the concentration of OH- ions is equal to the concentration of NH3.
Using the formula for the concentration of OH- ions in a basic solution, we have:
pOH = 14 - pH
pOH = 14 - 11.17
pOH = 2.83
Now, we can convert pOH to the concentration of OH- ions:
OH- concentration = 10^-pOH
OH- concentration = 10^(-2.83)
Assuming that the concentration of NH3 is the same as the concentration of OH- ions, we can calculate the molarity by using the OH- concentration:
Molarity = OH- concentration
Molarity = 10^(-2.83) M
So, the molarity of the NH3 solution is approximately 0.00595 M. You could say it's quite concentrated in terms of the NH3 concentration but relatively low in terms of the OH- concentration.
To calculate the molarity of a solution, we need to make use of the pH value and the pKa value of the solute.
In this case, NH3 (ammonia) acts as a weak base in water and undergoes partial ionization according to the following equation:
NH3 + H2O ⇌ NH4+ + OH-
The pKa value for the reaction is given as 4.75, which corresponds to the equilibrium constant expression:
pKa = -log10 (Kw / Ka), where Kw is the ion product of water (1.0 x 10^-14) and Ka is the acid dissociation constant of NH4+.
To determine the concentration (molarity) of the NH3 solution, we can start by finding the concentration of OH- ions using the pH:
pOH = 14 - pH = 14 - 11.17 = 2.83
Now, we can calculate the concentration of OH- ions:
[OH-] = 10^(-pOH)
= 10^(-2.83)
The concentration of NH4+ ions will be equal to the concentration of OH- ions, as both are formed in a 1:1 ratio. Therefore, we can substitute [OH-] with [NH4+].
Finally, to calculate the molarity, we need to know the volume of the solution. Let's assume a volume of 1 liter (L) for simplicity.
Now, the molarity (M) can be calculated as:
Molarity = moles solute / volume solvent
To find the moles of solute, we multiply the concentration ([NH4+]) by the volume (V) of the solution:
moles NH4+ = [NH4+] x V
= [OH-] x V
= 10^(-2.83) x 1
Therefore, the molarity of the NH3 solution, assuming a volume of 1 L, is approximately:
Molarity = Moles NH4+ / Volume (in L)
= (10^(-2.83) mol/L) / 1 L
Calculating this expression will provide the final molarity value for the NH3 solution.
To calculate the molarity of a solution, we need to know the volume of the solution and the number of moles of solute present.
In this case, we have a solution of NH3 in water, and we know the pH of the solution is 11.17. The pH of a solution is a measure of its acidity or basicity. In this case, a pH of 11.17 indicates that the solution is basic.
NH3 is a weak base, and in water, it can ionize to form NH4+ and OH-. The OH- ions contribute to the solution's basic nature.
To calculate the molarity, we can consider the concentration of OH- ions in the solution. The concentration of OH- ions can be calculated using the formula:
[OH-] = 10^(-pOH)
where pOH is the negative logarithm of the OH- concentration.
In this case, we have a pH of 11.17. To find the pOH, we can use the formula:
pOH = 14 - pH
Therefore, pOH = 14 - 11.17 = 2.83.
Now, we can calculate the OH- concentration:
[OH-] = 10^(-pOH) = 10^(-2.83)
To find the molarity, we need to know the volume of the solution. Let's assume we have 1 liter (L) of the solution.
Molarity (M) is defined as moles of solute per liter of solution. In this case, the solute is NH3.
Since NH3 is a weak base, it partially ionizes in water to form NH4+ and OH-. The reaction can be represented as:
NH3 + H2O --> NH4+ + OH-
The balanced equation shows that 1 mole of NH3 forms 1 mole of NH4+ and 1 mole of OH-. Therefore, the concentration of NH3 is equal to the concentration of OH- ions.
Hence, the molarity of the NH3 solution can be expressed as:
Molarity (M) = [OH-] = 10^(-2.83) = 0.00143 M
Therefore, the molarity of the NH3 solution is 0.00143 M.
pH = 11.17 so pOH = 14-11.17
Then pOH = -log(OH^-). Calculate OH^-
..........NH3 + H2O ==> NH4^+ + OH^-
Plug in OH^- (also equal to NH4^+) into Kb expression and solve for NH3.