Question

You decide to use formic acid 1.5L of 70mM buffer and you find the concentrationsto be 0.025M of formic acid and 0.045M of its conjugate base, you find that you can buy 19M formic acid from a chemical company.

a) what volume of 19M formic acid should you start with?

b) Would you make your desired buffer by adding strong acid (0.5 M HCl) or strong base ( 0.5M NaOH) why?

c) How many mL of either the strong acid or the strong base would you need to add?

Answer 1

Start from the question (b) first. You need to make a solution of weak acid HCOOH and its conjugate base HCOO^- to prepare an acidic buffer solution. You can get HCOO- by adding NaOH to HCOOH solution.

HCOOH + NaOH = HCOO^- + Na⁺ + H₂O

Think about the process. You first take a solution of HCOOH and add NaOH to it so that the final volume is 1.5 L and concentration of remaining acid becomes 0.025M and that of the salt produced is 0.045M. This resultant solution is the buffer solution you want to prepare.

So you need to find the initial molarity of HCOOH and how much of this acid solution you need.

Now we come to the calculation part.

Molarity of acid in final solution = 0.025 M

So 1.5 L buffer solution contains 0.025*1500/1000 or 0.0375 moles of acid.

Molarity of salt in final solution = 0.045 M

So 1.5 L buffer solution contains 0.045*1500/1000 or 0.0675 moles of the salt.

So we can say that initial solution of acid contains total (0.0375+0.0675) or 0.105 moles of acid. When you add 0.5M of NaOH solution, 0.0675 moles of acid react with NaOH to give 0.0675 moles of the salt and 0.0375 moles of the acid remains in the solution.

HCOOH + NaOH = HCOONa + H₂O

According to the reaction 1 mol of the acid react with 1 mol of NaOH. So 0.0675 moles of the base will react with 0.0675 moles of the acid.

We get, 0.0675*1000/0.5 mL or 135 mL of 0.5M NaOH contains 0.0675 moles of NaoH. (Answer of question c)

Come to the question (a). You need to prepare 1500-135 or 1365 mL of the acid solution which contains 0.105 moles of HCOOH. Molarity of this initial solution is 0.105*1000/1365 = 0.077 M

You have to prepare 1365 mL 0.077M HCOOH solution from 19M concentrated solution of HCOOH.

Volume needed of 19M HCOOH solution = 1365*0.077/19 = 5.53 mL (Answer of question a).