Question

You have to prepare a pH 5.00 buffer, and you have the following 0.10M solutions available: HCOOH, HCOONa, CH3COOH, CH3COONa, HCN, and NaCN.

A. The solutions you would see are CH3COOH, and CH3COONa...
B. How many millileters of each solution would you use to make approximately a liter of the buffer?

Can someone please help me with letter B? I don't know how to approach this problem.

Answer 1

@CausticSyndicalist do you know?

Answer 2

You can use \(\sf C_1V_1 = C_2V_2\)
Where C = concentration
and V = volume
the 1 represents initial (what you're starting with) and 2 represents the final volume/concentration you want.

Keep in mind that concentration is usually in units \(\sf \frac{mol}{L}\)

Answer 3

There is no starting volume listed...how would we do that then

Answer 4

@abb0t

Answer 5

Start with the Henderson Hasselbalch equation: \(pH=pKa+log\dfrac{[A^-]}{[HA]}\)

Answer 6

ok so far i have \[5.00=4.74 +\log(CH _{3}COO ^{-}/CH _{3}COOH)\]

Answer 7

then you would subtract the 4.74 over and get \[.26=\log(CH _{3}COO ^{-}/CH _{3}COOH)\] and to get rid of the log you do \[10^{.26}\] which equals \[1.82\] then you will have \[1.82=(CH _{3}COO ^{-}/CH _{3}COOH)\] and this is where i'm stuck

Answer 8

\(\dfrac{[A^-]}{[HA]}=\dfrac{1.82}{1} \)

the total moles (assuming 1 L) are: \(M=\dfrac{n}{L}\rightarrow n=0.1~M*1L=0.1`moles\)

so in terms of percent:

\([A^-]=\dfrac{1.82}{1+1.82}*0.1~moles\)

\([HA]=\dfrac{1}{1+1.82}*0.1~moles \)

Find the volume representative of those moles

Answer 9

A- would be .1
and HA would be .035 or 3.5*10^-2 ???

Answer 10

moles of \(A^-\) are 0.64539 moles, therefore

\(\sf 0.1~M=\dfrac{0.64539~moles}{L}\rightarrow L=0.6454~L ~or ~645.4 ~mL \)

Answer 11

sorry it was 0.064539 moles

Answer 12

ok but what would HA be then? sorry, i'm just pretty confused on this

Answer 13

from above, the moles of HA are:

\(\sf n_{HA}=\dfrac{1}{1+1.82}*0.1~moles=0.03546~moles\)


then the volume is :

\(\sf 0.1~M=\dfrac{0.03546~moles}{L}\rightarrow L=0.3546~L~or~345.6~mL\)