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Calculate the molar ratio of NaF to HF required to create a buffer with pH=4.20. Ka(HF)=6.8x
10^(-4)

User Didil
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To calculate the molar ratio of NaF to HF required to create a buffer with pH=4.20, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the dissociation constant (Ka) of the weak acid (HF) and the ratio of the concentrations of the weak acid and its conjugate base (F-):

pH = pKa + log([F-]/[HF])

We know the pH and Ka of the buffer, so we can rearrange the equation to solve for [F-]/[HF]:

[F-]/[HF] = 10^(pH - pKa)

[F-]/[HF] = 10^(4.20 - 6.8) = 0.005012

The molar ratio of NaF to HF required to create this buffer depends on the concentration of the buffer solution, which is not specified in the question. However, we can assume that the buffer is made with equimolar concentrations of NaF and HF, in which case the ratio of their molar amounts would be:

mol NaF : mol HF = [F-] x Vol : [HF] x Vol

where Vol is the total volume of the buffer solution in liters.

Since we assume equimolar concentrations of NaF and HF, we can set [F-] equal to [HF]:

[F-] = [HF]

Therefore:

mol NaF : mol HF = [F-] x Vol : [HF] x Vol = 0.005012 x Vol : 1 x Vol

mol NaF : mol HF = 0.005012 : 1

mol NaF : mol HF = 1 : 199.2

Therefore, the molar ratio of NaF to HF required to create a buffer with pH=4.20 and Ka(HF)=6.8x10^-4 is approximately 1:199.2.

User Alex Blokha
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