A. if you prepare 250.0mL of pH=3.60 buffer that has a total buffer concentratio

Question

A. if you prepare 250.0mL of pH=3.60 buffer that has a total buffer concentration of formic acid+ formate of 0.030 M, how many moles of each will you need to prepare the solution in question 1? question 1:Determine the number of moles of reagent in the following solutions: 1a)25.00 mL of 0.10 M acetic acid =0.0025 moles 1b) 5.55 mL of 0.092 M NaOH =0.5106 moles 1c)0.50 ml of 0.087 M HCL=0.435 moles B. discuss how to prepare the solution in question A starting from 0.100 M formic acid and 0.150 M sodium formate. PLEASE SHOW ALL WORK

Explanation / Answer

1))

ph=pka+log(salt concentration/acid concentration)
3.6=3.77+log(x/.03-x)
=>x=.012
therefore concentration of salt and acid are .012 and .018
moles =concentration*1000/volume
moles of salt =.012*1000/250 here to prepare 250 ml solution
=.048 moles
moles of acid= .018*4
=.072moles

2)

molarity = moles/volume in litres

moles=molarity *Volume(in L)

a)

ch3cooh = 0.1 *(25*10^-3)=0.0025 moles...

similrly b and c.

3)

ph=pka+log(salt concentration/acid concentration)

to prepare required solution starting from 0.100 M formic acid and 0.150 M sodium formate

salt conc = [formate]=0.012M

acid conc =[formic acid]=0.108M

final volume must be 250 ml.

moles of salt=0.048=0.1*V1

V1=48 ml of 0.1M acid sloution

moles of acid =0.072=0.150*V2

V2=48 ml of 0.150M formate solution

take 48 ml of 0.1M formic acid sloution and add it to 48 ml of 0.150M formate solution and dilute the solution to 250 ml.

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