# I have a titration question dealing with biochemistry

• Workout
When the solution is 1/4 titrated, there will be a 4 fold change in the concentration of the buffer and acid. Apply this relationship to your problem.In summary, you are titrating a 75 mL volume of 33 mM lactate buffer (pKa 3.86) with 1.3 M HCl and adding 1 mL of the HCl. The concept of deprotonated and fully protonated refers to the presence or absence of H+ ions in the buffer solution. To solve this problem, you will need to calculate the moles of buffer and H+ ions, use the Henderson-Hasselbalch equation, and consider the relationship between buffer pH and concentration during titration.

## Homework Statement

You are titrating a 75 mL volume of 33 mM lactate buffer (pKa 3.86) with 1.3 M HCl (starting fully deprotonated). What is the pH after you have added 1 mL of the HCl? Give 2 decimal places

## Homework Equations

pH = pKa + log[A-/HA]

## The Attempt at a Solution

I don't really understand the concept of deprotonated and fully protonated. Could someone explain the two? I also have no idea where to even start with this question...

Deprotonated buffer - No H+ ions present in the buffer solution, meaning the buffer can no longer digest the base.

If you add H+ in the buffer, you are protonating it, thus you are decreasing the pH of the buffer. If you want to attempt the question, ask yourself the following questions:-

1. How many moles of buffer I've been given? How much moles of H+ I am adding?

2. How much H+ can 1 mole of lactate buffer can assimilate?

3. Can I use the relevant equation?

You are not "protonating a buffer". You are protonating the conjugate base. So if there is only acid from (HLactate) in the solution, lactate is fully protonated. That will happen at low pH. If there is a mixture of HLactate and Lactate- - it is protonated only partially, that happens when the pH is relatively close to pKa. Finally, in high pH all lactate is present in Lactate- form and it is not protonated at all.

You can assume protonation is quantitative - all strong acid added reacts with Lactate- yielding HLactate. Use this information to calculate new concentrations of acid and conjugate form and plug them into the Henderson-Hasselbalch equation.

Also compare this buffer pH change problem.