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???
4 answers
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)
.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
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.
1 answer