question is in the description

Question

A 28.9 mL sample of 0.327 M ammonia, NH₃, is titrated with 0.323 M hydrochloric acid.

After adding 12.3 mL of hydrochloric acid, the pH is .

Use the Tables link in the References for any equilibrium constants that are required.

Answer 1

Let's look at the equation for the reaction taking place:

NH3 + HCl ==> NH4Cl

This makes a BUFFER because you have a weak base (NH3) and the conjugate acid (NH4⁺)

To find the pH or pOH of a buffer solution, we can use the Henderson Hasselbalch equation which for a basic buffer is

pOH = pKb + log [salt]/[base]

So, we must first find the concentration of the buffer components:

moles NH3 = 0.0289 L x 0.327 mol/L = 0.00945 moles NH3

moles HCl = 0.0123 L x 0.323 mol/L = 0.00397 moles HCl

NH3 + HCl ==> NH4Cl

0.00945..0.00397.....0.......Initial

-0.00397..-0.00397....+0.00397..Change

0.00548......0..............0.00397...Equilibrium

Final Volume = 28.9 ml + 12.3 ml = 41.2 ml = 0.0412 L

Final [NH3] = 0.00548 mol/0.0412 L = 0.133 M

Final [NH4+] = 0.00397/0.0412 L = 0.0964 M

pOH = pKb + log [salt]/[base] (looking up pKb for NH3 I find it to be 4.75

pOH = 4.75 + log (0.0964/0.133) = 4.75 + log 0.725

pOH = 4.75 + (-0.14)

pOH = 4.61

pH = 14 - pOH

pH = 9.39