Q19.107CP - Chemistry: Molecular Nature Of Matter And Change

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  $\mathbf{\text{1.0 M HCOOH}}$and  data-custom-editor="chemistry" $\mathbf{\text{1.0 M NaOH}}$? 

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

Step-by-Step Solution

Answer

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

(b) (b)Solution from stock solutions  $\text{is 0.4676L.}$

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

Step 1: 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  .$\frac{\left[{\text{A}}^{\text{-}}\right]}{\text{[HA]}}\text{=0.989}$

Step 3: 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 $\begin{array}{c}\text{x=0.2324L=V(KOH)}\\ \text{V(HCOOH)=0.7L-0.2324L}\\ \text{=0.4676L}\end{array}$

Step 4: 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  $\begin{array}{c}\text{[HCOOH]=}\frac{\text{0.4676L×1M-0.2324L×1M}}{\text{0.7L}}\\ \text{=0.3359M}\end{array}$