Question

As an FDA physiologist, you need 0.700 L of formic acid–formate buffer with a pH of 3.74.

(a) What is the required buffer-component concentration ratio?

(b) How do you prepare this solution from stock solutions of $\mathbf{\text{1.0MHCOOH}}$and data-custom-editor="chemistry" $\mathbf{\text{1.0MNaOH}}$?

(c) What is the final concentration of data-custom-editor="chemistry" $\mathbf{\text{HCOOH}}$in this solution?

Short answer

(a) The required buffer-component concentration ratio is $\text{0.989.}$

(b) (b)Solution from stock solutions $\text{is0.4676L.}$

(c) (c)The final concentration is $\text{0.3359M}.$

Step-by-step solution

Concept Introduction

In polar liquids, the ionic material dissociates into its ions in ionic equilibrium. The ions generated in the solution are always in equilibrium with the undissociated solute.

Step 2:What is the required buffer-component concentration ratio?

Simplify the values:

$\mathrm{V}$(formic acid-formate buffer)=0.7L

pH( formic acid-formate buffer )=3.74

c(HCOOH)=1 M

c(NaOH)=1M

a) The acid-base ratio can be calculated using the Henderson-Hasselbalch equation:

${\text{Ka(HCOOH)=1.8×10}}^{\text{-4}}$

$\text{pH=pKa+log}\frac{\left[{\text{A}}^{\text{-}}\right]}{\text{[HA]}}$

$\text{3.74=-log}\left({\text{1.8×10}}^{\text{-4}}\right)\text{+log}\frac{\left[{\text{A}}^{\text{-}}\right]}{\text{[HA]}}$

Therefore, the required buffer-component concentration ratio .$\frac{\left[{\text{A}}^{\text{-}}\right]}{\text{[HA]}}\text{=0.989}$

 How do you prepare this solution from stock solutions?

b) We must keep in mind that we are preparing a 0.7L solution for this problem.

$\text{0.989=}\frac{\text{xL×1M}}{\text{(0.7L-x)×1M-xL×1M}}$

We receive this V value when we conclude our mathematic calculations:

Therefore, solution from stock solutions$\begin{array}{c}\text{x=0.2324L=V(KOH)}\\ \text{V(HCOOH)=0.7L-0.2324L}\\ \text{=0.4676L}\end{array}$

 What is the final concentration of HCOOH in this solution?

c) To compute the [HCOOH], we just use the information we calculated in the preceding problems:

Therefore,the final concentration of $\begin{array}{c}\text{[HCOOH]=}\frac{\text{0.4676L×1M-0.2324L×1M}}{\text{0.7L}}\\ \text{=0.3359M}\end{array}$