Ammonia (NH3) ionizes according to the following reaction:
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)
The base dissociation constant for ammonia (NH3) is Kb = 1.8 × 10–5. Ammonia (NH3) also has a chloride salt, ammonium chloride (NH4Cl), which is soluble in water.
If 0.070 M of ammonia (NH3) and 0.035 M of its salt ammonium chloride (NH4Cl) are mixed in a solution, what is the pH of this solution?
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To find the pH of the solution, we need to determine the concentration of the hydroxide ion (OH-) and then calculate the pOH, which can be converted to pH using the equation pH = 14 - pOH.
First, let's calculate the concentration of OH- in the solution.
In the given reaction, NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq), we know that for every 1 mole of NH3 (ammonia) that ionizes, 1 mole of OH- is produced.
Given that the concentration of NH3 is 0.070 M, we can assume that the concentration of OH- is also 0.070 M.
Next, we need to calculate the concentration of OH- resulting from the dissociation of NH4Cl.
NH4Cl (ammonium chloride) ionizes according to the equation NH4Cl(s) ⇌ NH4+(aq) + Cl-(aq).
Since ammonium chloride is soluble, it will completely dissociate in water, yielding NH4+ and Cl- ions in equal concentrations.
Given that the concentration of NH4Cl is 0.035 M, the concentration of NH4+ and Cl- is also 0.035 M.
Therefore, the concentration of OH- due to the dissociation of NH4Cl is zero since there are no OH- ions produced.
Now, we can calculate the total concentration of OH- in the solution:
Total OH- concentration = OH- from NH3 + OH- from NH4Cl
Total OH- concentration = 0.070 M + 0 M
Total OH- concentration = 0.070 M
Now, let's calculate the pOH:
pOH = -log10(OH- concentration)
pOH = -log10(0.070)
pOH ≈ 1.15
Finally, we can find the pH using the equation pH = 14 - pOH:
pH = 14 - 1.15
pH ≈ 12.85
Therefore, the pH of the solution is approximately 12.85.