Formic acid (HA) has a Ka value of 1.8x10-4 . A student needs to prepare 20.0 mL

Question

Formic acid (HA) has a Ka value of 1.8x10-4 . A student needs to prepare 20.0 mL of a buffer with pH=3.60 from stock solutions of 0.10 M HA and 0.10 M NaA (where “A” represents the formate ion). What volumes of the 0.10 M HA solution and the 0.10 M NaA solution should be mixed to prepare the buffer? (Hint: See the example in the section on “Preparing a Buffer”. First use the pH to calculate [H3O + ]. Then use equation 5 to calculate the value of the ratio [HA]o/[ A]o . Then let X = the number of mL of 0.10 M HA to use. How many mL of the NaA would then be needed? Use C1V1=C2V2 to find equations for the initial concentrations of formic acid [HA]o and its conjugate base [A]o after mixing, but before the ionization reaction. These equations will involve the unknown value of X. Then divide one by the other and solve for X by using the value of the [HA]o/[ A]o ratio that you first calculated.)

Explanation / Answer

Ka = 1.8*10^-4

HA <--> H+ and A-

will be the equilibrium

K = [H+][A-]/[HA]

You can represent this in the pH equation os Henderson Hasslebach, which is for buffers only

pH = pKa + log(A-/HA)

if you need pH = 3.6

calculate pKa = -log(Ka) = -log(1.8*10^-4 ) = 3.75

Now you can find the ratio of acid/conjugate

pH = pKa + log(A-/HA)

3.6 = 3.75 = log(A-/HA)

(A-/HA) = 0.7079

A- = 0.7079*HA

since M1 = M2

we can assume

1 ml of Ha per 0.7079 ml of A-

Total volume = 1+0.7079 = 1.7079

but we need a total of 20 ml so:

20/1.7079 = 11.71

Therefre..

1*11.71 + 0.7079*11.71 = 20

Volume of acid = 11.71 ml

Volume of NaAc = 8.29 ml

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