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:

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

c (diamond),

sio2 (quartz)