A Student is going to titrate a 25.0 mL sample of 0.150 M hydrofluoric acid with

Question

A Student is going to titrate a 25.0 mL sample of 0.150 M hydrofluoric acid with a 0.150 M NaOH solution. What is the pH at the equivalence point? The Ka of hydrofluoric acid is 3.5x10^-4

6) The initial pH of HF

Table A, B, C, D, or E?

7) What is the next step to the problem above?

Take log x to solve for pOH and then subtract 14

Nothing the problem is done

Put in Henderson-Hasselbalch equation

Plug into the Kb expression, solbe for x

Plug into the Ka expression, solve for x?

This is an assignment that is worth a huge amount of points. I have about half of it figured out but I am completely stuck on the titration part.

Explanation / Answer

HF + NaOH =NaF + H2O

Here HF is weak acid and NaF is the conjugate base.

As both HF and NaOH have same molarity, it will take 25 mL of NaOH to reach equivalence point.

After equivaence point, there will be zero moles of NaOH and zero moles of HF and 7.50E-3 mole of NaF

Again, Kb=Kw/Ka thus Kb=1.0E-14/3.50E-4 = 2.85E-11

Moles of F- = 7.50E-3
Total volume = 50.00 mL

So, molarity of F- = (7.50E-3 mole)/(0.050L) = 1.5E-1M

Now using the ICE table we come up with the following equation

Kb=[HF][OH-]/[F-] and since the Kb value is so low we can ignore the x on the denominator leaving us with the follwing equation

2.85E-11=(x*x)/0.150M

or x = sqrt(2.85E-11*0.150) = 2.07E-6 = [OH-]

So, pOH = -log[OH-] = -log(2.07E-6) = 5.68

Thus, pH = 14.00 - 5.68 = 8.32

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