Question

0/2 points Previous Answers ZumChemP8 8.E.042. My Notes Ask Your Teacher What volumes of 0.53 M HF and 0.53 M NaF must be mixed to prepare 1.00 L of a solution buffered at pH = 2.79?

Answer 1

V(NaF) = 0.309 L, V(HF) = 0.691 L

Step-by-step explanation:

The pH of the given buffer solution is calculated using:

pH = pka + log {[F-]/[HF]}

for HF, pka=3.14

Therefore,

2.79 = 3.14 + log {[F-]/[HF]}

log {[F-]/[HF]}=2.79-3.14=-0.35

{[F-]/[HF]} = 10^(-0.35)=0.447

Assuming that NaF dissociates completely, we have:

[F-]= [HF]

The concentration of NaF and HF in the buffer solution are related to the concentration of the initial solutions [F-] and [HF] as shown below:

[F-] = {[F-]o*V(NaF)}/V(bs) = 0.53*V(NaF)/1=0.53V(NaF)

Similarly,

[HF] = {[HF]o*V(HF)}/V(bs) = 0.53*V(HF)/1=0.53V(HF)

Where:

V(NaF) and V(HF) are the volumes of the initial solutions. V(bs) is the solution of the resulting buffer solution.

We also know that:

V(NaF) + V(HF) = V(bs) = 1

[F-]/[HF] = 0.447

If V(NaF) = x, V(HF) = y

Then,

x+y =1: x/y=0.447

x=0.447*y

0.447y+y=1

y = 1/1.447 = 0.691

x = 1-y = 1-0.691 = 0.309

Therefore, V(NaF) = 0.309 L V(HF) = 0.691 L

Answer 2

304 mL of NaF and 696 mL of HF

Step-by-step explanation:

A buffer is a solution of a weak acid, or a weak base, in equilibrium with its conjugate base, or conjugate acid. Because of the equilibrium, the buffer can control the pH of the solution, which remains almost unaltered bu the addition of a base or an acid.

In this case, the equilibrium is:

HF ⇄ H⁺ + F⁻

And F⁻ comes from the salt NaF, which dissociates completely, so [NaF] = [F⁻].

The pH of a buffer can be calculated by the Handerson-Halsebach equation:

pH = pKa + log [A⁻]/[HA]

Where pKa = -logKa, Ka is the constant of the equilibrium of the acid, [A⁻] is the concentration of the conjugate base, and [HA] is the concentration of the acid. For HF, pKa = 3.15. So:

2.79 = 3.15 + log [F⁻]/[HF]

log [F⁻]/[HF] = -0.36

[F⁻]/[HF] = 0.4365 (equation 1)

These concentrations are at the 1.00L buffer. The volume of the dilution can be calculated by:

C1V1 = C2V2

Where C is the concentration, V is the volume, 1 is the concentrated solution, and 2 is the diluted solution.

For HF:

0.53*VHF = [HF]*1

[HF] = 0.53*VHF (equation 2)

For F⁻:

0.53*VNaF = [F⁻]*1

[F⁻] = 0.53*VNaF (equation 3)

And, VHF + VNaF = 1.00

VHF = 1 - VNaF (equation 4)

Substituing equation 4 in equation 2:

[HF] = 0.53*(1 - VNaF) (equation 5)

Substituting equations 3 and 5 in equation 1:

[0.53*VNaF]/[0.53*(1 - VNaF)] = 0.4365

VNaF/(1 - VNaF) = 0.4365

VNaF = 0.4365 - 0.4365*VNaF

1.4365*VNaF = 0.4365

VNaF = 0.3040 L = 304 mL

VHF = 1 - 0.3040

VHF = 0.6960 L = 696 mL