Calculate the pH of the solution that results from the addition of 12.00mL of 0.225M HNO3 to 20.00mL of 0.210M NH3. Buffer titration problem.

Do I find moles and then???

Find moles, work up an ICE chart to determine which reagent (HNO3 or NH3 is in excess), then apply the Henderson-Hasselbalch equation to find the pH.

.0027moles HNO3 - .0027moles HNO3

.0042moles NH3 - .0027 moles = .0015moles NH3 in excess. Therefore is this correct?

pH = pKa + log(x^2)/(.0027)(.0015)

moles HNO3 correct.

moles NH3 correct.
ICE chart must be ok since moles NH3 in excess ok. But at that point you lost me.
What's with X^2.
pH = pKa + log[(base)/(acid)
pH = ??
pKa = about 9.24 but you need to look up Kb in your text, convert to pKb, then subtract from 14 for pKa.
Then for base in the HH equation you substitute moles NH4NO3 (0.00150)and divide by volume. For acid substitute 0.00270 divided by volume. or since volume cancels you can simply substitute moles. Without V it looks this way.
pH = 9.24 + log(0.0015/0.00270) = about 8.98

Thank you so much. I see now. You are more helpful than the professor we have now. I appreciate the time you give to this site.

To calculate the pH of the solution resulting from the addition of HNO3 to NH3, we need to consider the reaction between the two.

NH3(aq) + HNO3(aq) → NH4NO3(aq)

This reaction produces ammonium nitrate (NH4NO3), which is a salt. The NH4+ ion from the salt acts as the conjugate acid, and the NO3- ion acts as the conjugate base. The NH3 (ammonia) acts as the weak base, and HNO3 (nitric acid) acts as the strong acid.

In a buffer solution, the equilibrium between the weak base and its conjugate acid helps maintain the pH by resisting changes in acidity or basicity when small amounts of acid or base are added to the solution.

To solve this buffer titration problem, follow these steps:

1. Calculate the moles of HNO3:
Moles = volume (L) × molarity (M)
Moles of HNO3 = 12.00 mL (0.01200 L) × 0.225 M = 0.00270 moles

2. Calculate the moles of NH3:
Moles = volume (L) × molarity (M)
Moles of NH3 = 20.00 mL (0.02000 L) × 0.210 M = 0.00420 moles

3. Calculate the moles of NH4+ and NO3- formed (from the reaction between NH3 and HNO3):
The ratio of NH3 to NH4+ is 1:1, and the ratio of HNO3 to NO3- is also 1:1.
The moles of NH4+ and NO3- formed will be equal to the number of moles of HNO3 used.

Moles of NH4+ = 0.00270 moles
Moles of NO3- = 0.00270 moles

4. Calculate the concentrations of NH4+ and NO3-:
Concentration (M) = moles / volume (L)

Concentration of NH4+ = 0.00270 moles / (0.01200 L + 0.02000 L) = 0.0675 M
Concentration of NO3- = 0.00270 moles / (0.01200 L + 0.02000 L) = 0.0675 M

5. Determine the composition of the resulting solution:
The resulting solution will contain the original NH3 and NH4+ ions, as well as the NO3- ions from the HNO3 added.

6. Calculate the pH of the solution using the Henderson-Hasselbalch equation:
pH = pKa + log ([conjugate base] / [weak acid])

In this case, NH3 acts as the weak base and NH4+ acts as the conjugate acid.
The pKa value for the ammonium ion (NH4+) is typically around 9.25 at 25°C.

pH = 9.25 + log (0.0675 M / 0.0675 M)
pH = 9.25 + log (1)
pH = 9.25

Therefore, the pH of the solution resulting from the addition of 12.00 mL of 0.225 M HNO3 to 20.00 mL of 0.210 M NH3 is 9.25.