Question #99405

1 Answer
Stefan V.
Apr 12, 2015

!! VERY LONG ANSWER !!

Since you're dealing with a buffer solution, you can use the Henderson-Hasselbalch equation to solve for the pH of the solution.

Now, since you mistyped the volume of the solution, I'll assume it to be 0.100 L - you can use whatever value you had.

So, according to the Henderson-Hasselbalch equation,

pHsol=pKa+log([F−][HF]), (1)

You also need the acid dissociation constant for HF, which is listed as being Ka=6.7⋅10−4.

The pKa will be

pKa=−log(Ka)=−log(6.7⋅10−4)=3.17

Plug your values into equation (1) to get the pH

pHsol=3.17+log(0.50 M0.25 M)=3.17+log(2)=3.47

Now you start adding stuff to your buffer. When you add nitric acid, HNO3, which is a strong acid, it will react with the conjugate base of your weak acid, F−.

As a result, you'll have less NaF in the solution, which implies that more HF will be produced.

HNO3(aq)+NaF(aq)→HF(aq)+NaNO3(aq)

Now, you need to calculate how many moles of NaF and of HF you initially had in the buffer. To do this, use the given volume and molarity

C=nV⇒n=C⋅V

nNaF=0.50 M⋅0.100 L=0.050 moles NaF

nHF=0.25 M⋅0.100 L=0.025 moles HF

Set up your ICE table

HNO3(aq)+NaF(aq)→HF(aq)+NaNO3(aq)
I.......0.002...............0.05..............0.025
C.....(-0.002)...........(-0.002).......(+0.002)
E.......0......................0.048............0.027

All the moles of HNO3 will be consumed; at the same time, 0.002 moles of HF will be produced and the number of moles of NaF will decrease by 0.002.

Calculate the new molarities of the species present in the buffer

[HF]=0.027 moles0.100 L=0.27 M

[F−]=0.048 moles0.100 L=0.48 M

Use equation (1) to determine the new pH

pHsol 2=3.17+log(0.48 M0.27 M)=3.42

Adding a strong acid to your buffer reduced the pH slightly

Now you add potassium hydroxide, KOH - a strong base. It will react with the weak acid to produce F−.

KOH(aq)+HF(aq)→KF(aq)+H2O(l)

SIDE NOTE You can ignore the K+ cation altogether, the important thing in this reaction is F−.

Once again, set up the ICE table

OH−(aq)+HF(aq)→F−(aq)+H2O(l)
I......0.004...........0.027............0.048
C...(-0.004).......(-0.004).........(+0.004)
E.........0...............0.023..............0.052

[HF]=0.023 moles0.100 L=0.23 M

[F−]=0.052 moles0.100 L=0.52 M

Finally, use (1) to solve for the new pH

pHsol 3=3.17+(0.52 M0.23 M)=3.52

Adding a strong base increased the pH slightly.

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