A chemist needs to prepare a buffer solution of pH 8.80. What molarity of NH3 (pKb = 4.75) is required to produce the buffer solution if the (NH4)2SO4 in the solution is 1.8 M?

The answer I got was 1.125 moles

To prepare a buffer solution of pH 8.80, we need to use the Henderson-Hasselbalch equation:

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

Here, NH4+ (from (NH4)2SO4) acts as the conjugate acid (HA) and NH3 acts as the conjugate base (A-).

Given information:
pKb (dissociation constant of NH3) = 4.75
(NH4)2SO4 concentration = 1.8 M

Step 1: Calculate pKa
pKa = 14 - pKb
pKa = 14 - 4.75
pKa = 9.25

Step 2: Calculate NH3 concentration ([A-])
[A-] = [NH4+] = 1.8 M (concentration of (NH4)2SO4)

Step 3: Calculate HA concentration ([NH4+])
[HA] = 10^(pH - pKa) * [A-]
[HA] = 10^(8.80 - 9.25) * 1.8
[HA] = 10^(-0.45) * 1.8
[HA] = 0.33 * 1.8
[HA] = 0.594 M

So, the required concentration of NH3 (A-) to prepare the buffer solution is approximately 0.594 M.

To prepare a buffer solution of pH 8.80 using the Henderson-Hasselbalch equation, we need to calculate the ratio of the concentration of the conjugate base (NH3) to that of the weak acid (NH4+).

The Henderson-Hasselbalch equation is given by:

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

Here, the weak acid is NH4+ (formed from (NH4)2SO4), and the conjugate base is NH3.

First, let's calculate the pKa from the pKb of NH3:

pKw = pKa + pKb

Since pKw = 14 (at 25°C), we can rearrange the equation to find pKa:

pKa = pKw - pKb
= 14 - 4.75
= 9.25

Now, let's use the Henderson-Hasselbalch equation to find the ratio of [NH3] to [NH4+] required to obtain a pH of 8.80:

8.80 = 9.25 + log([NH3]/[NH4+])

Rearranging the equation:

log([NH3]/[NH4+]) = 8.80 - 9.25
= -0.45

Using the properties of logarithms, we can raise both sides of the equation as powers of 10:

[NH3]/[NH4+] = 10^(-0.45)
= 0.35

Thus, the ratio of [NH3] to [NH4+] required is 0.35. Since we know the concentration of NH4+ in the solution is 1.8 M, we can multiply this concentration by the ratio to find the concentration of NH3:

[NH3] = 0.35 * 1.8 M
= 0.63 M

Therefore, a molarity of 0.63 M of NH3 is required to prepare the buffer solution of pH 8.80 when the (NH4)2SO4 concentration is 1.8 M.

Use the HH equatiion. (acid) = 2*1.8 and x = (base)