How many moles of sodium hydroxide would have to be added to 250 mL of a 0.303 M hydrofluoric acid solution, in order to prepare a buffer with a pH of 3.630?

0.0562 moles

Explanation:

Based on the H-H equation, for the HF buffer:

pH = pKa + log [F⁻] / [HF]

Where pH is the pH of the buffer = 3.630

pKa is pKa of HF buffer: 3.17

And [] could be taken as the moles of each reactant

3.630 = 3.17 + log [F⁻] / [HF]

0.46 = log [F⁻] / [HF]

2.884 = [F⁻] / [HF] (1)

The initial moles of HF are:

0.250L * (0.303mol / L) = 0.07575 moles

The HF reacts with NaOH as follows:

HF + NaOH → NaF + H₂O

That means the moles of HF will be the initial moles of HF - Moles NaOH

And moles NaF = F⁻ = Moles of NaOH added

That means:

0.07575 moles = [F⁻] +[HF] (2)

Replacing (2) in (1):

2.884 = [F⁻] / 0.07575 moles - [F⁻]

0.218 - 2.884[F⁻] = [F⁻]

0.218 = 3.884[F⁻]

[F⁻]  = 0.0562 moles

That means the moles of NaOH that must be added are:

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

i believe it is c.) 1575 j.