How many grams of dry NH4Cl need to be added to 2.00L of a 0.100M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.86? Kb for ammonia is 1.8×10−5.

To solve this problem, we need to understand how a buffer solution works and how the pH of a solution can be calculated.

A buffer solution is a solution that resists changes in pH when a small amount of acid or base is added to it. It typically contains a weak acid and its conjugate base or a weak base and its conjugate acid. In this case, we have a weak base, ammonia (NH3), and its conjugate acid, ammonium chloride (NH4Cl).

The pH of a solution can be calculated using the equation: pH = pKa + log([A-]/[HA]), where pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

In this problem, we are given the desired pH (8.86) and the concentration of ammonia (0.100M). We need to calculate the amount of ammonium chloride (NH4Cl) to add to achieve this pH.

First, we need to find the pKa value for ammonia (NH3). Pka = -log(Ka), and since Kb + Ka = Kw, we can use Kb (1.8×10−5) to calculate Ka.

Ka = Kw/Kb = 1.0×10^(-14) / (1.8×10−5) = 5.56×10^(-10)

Now, let's substitute the known values into the pH equation:

8.86 = pKa + log([A-]/[HA])

We know that when the pH equals pKa, the ratio of [A-]/[HA] is 1. So, we can rewrite the equation as:

8.86 = log(1)

Now, let's solve for pKa:

pKa = 8.86

Now, we need to calculate the concentration of the ammonium ion ([NH4+]). Since ammonia (NH3) is a weak base, it reacts with water to form a small amount of ammonium ions (NH4+) and hydroxide ions (OH-). The balanced equation is:

NH3 + H2O ⇌ NH4+ + OH-

Given that the concentration of ammonia (NH3) is 0.100M, we can assume that the concentration of [NH4+] is also 0.100M in this case.

Now we need to calculate the concentration of NH3. NH4Cl dissociates completely in water:

NH4Cl → NH4+ + Cl-

Since the concentration of NH4+ is 0.100M, the concentration of NH4Cl is also 0.100M. Since there is a 1:1 ratio between the concentrations of NH3 (weak base) and NH4+ (conjugate acid) in the buffer solution, the concentration of NH3 is also 0.100M.

Now that we know the concentrations of NH4+ and NH3, we can calculate the ratio [A-]/[HA]:

[A-]/[HA] = [NH3]/[NH4+] = 0.100 / 0.100 = 1

Now, let's substitute the values into the pH equation:

pH = 8.86 = pKa + log(1)

This gives us:

pKa = 8.86

Since the desired pH and pKa are equal, no additional NH4Cl needs to be added.

In conclusion, no grams of dry NH4Cl need to be added to the 2.00L of the 0.100M ammonia solution to prepare a buffer solution with a pH of 8.86.

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