pH buffer question help (1 Viewer)

[MyHeadee]

What is the pH change of the following buffer solution?

25mL buffer solution which has equimolar concentrations (0.1011M) of acetic acid and sodium acetate. What will the pH be if I add 5mL of 0.1032M NaOH to this?

Thanks

 

[bigwilly69]

bump

 

[thush@decode]

Henderson-Hasselbach equation:

pH = pKa + log_10 ([Ac-]/[HAc]) = pKa + log_10 (n(Ac-)/n(HAc))

Given we are only looking for the change in pH, we just need to see how "log_10 (n(Ac-)/n(HAc))" changes since pKa is constant.

Initially, log_10 (n(Ac-)/n(HAc)) = 0, as the acid and conjugate base are in equal concentrations and amounts initially.

Now we just need to find what the new value of the fraction n(Ac-)/n(HAc) would be.

Initially, n(HAc) = n(Ac-) = 0.025 x 0.1011 = 0.0025275 mol (yes I am being sloppy with sig figs)

When you add in NaOH, the OH- will pull the protons off the HAc to form Ac-.
Now, n(NaOH) = 0.005 x 0.1032 = 0.000516 mol.
So, 0.000516 mol of HAc will be converted to Ac-.
Hence

• n(HAc)_new = 0.0025275 - 0.000516 = 0.0020115 mol
• n(Ac-)_new = 0.0025275 + 0.000516 = 0.0030435 mol
• log_10 (n(Ac-)/n(HAc)) = log_10 (0.0030435/0.0020115) = 0.180

Therefore, log_10 (n(Ac-)/n(HAc)) increases from 0 to 0.180 - increases by 0.180.

This means that pKa + log_10 (n(Ac-)/n(HAc)) will have increased by 0.180 (as pKa is constant).

Hence, pH change would be +0.180 (ignoring sig fig rules).

 

[bigwilly69]

Henderson-Hasselbach equation:

pH = pKa + log_10 ([Ac-]/[HAc]) = pKa + log_10 (n(Ac-)/n(HAc))

Given we are only looking for the change in pH, we just need to see how "log_10 (n(Ac-)/n(HAc))" changes since pKa is constant.

Initially, log_10 (n(Ac-)/n(HAc)) = 0, as the acid and conjugate base are in equal concentrations and amounts initially.

Now we just need to find what the new value of the fraction n(Ac-)/n(HAc) would be.

Initially, n(HAc) = n(Ac-) = 0.025 x 0.1011 = 0.0025275 mol (yes I am being sloppy with sig figs)

When you add in NaOH, the OH- will pull the protons off the HAc to form Ac-.
Now, n(NaOH) = 0.005 x 0.1032 = 0.000516 mol.
So, 0.000516 mol of HAc will be converted to Ac-.
Hence

• n(HAc)_new = 0.0025275 - 0.000516 = 0.0020115 mol
• n(Ac-)_new = 0.0025275 + 0.000516 = 0.0030435 mol
• log_10 (n(Ac-)/n(HAc)) = log_10 (0.0030435/0.0020115) = 0.180

Therefore, log_10 (n(Ac-)/n(HAc)) increases from 0 to 0.180 - increases by 0.180.

This means that pKa + log_10 (n(Ac-)/n(HAc)) will have increased by 0.180 (as pKa is constant).

Hence, pH change would be +0.180 (ignoring sig fig rules).

you're a gentleman and a scholar, thank you

 

[Qeru]

The Henderson-Hasselbach Equation isn't in the syllabus so a derivation is probably required. It's not too hard to prove:



The is defined as:

 

[CM_Tutor]

You can also skip the HH equation completely by solving the problem using an ICE table.