Adding a strong acid to buffer solution

[kevvyroo]

Homework Statement

I'm trying to find the pH change of adding 0.001 mol of HNO$_{}3$ to a buffer solution containing 0.24 M HF and 0.55 M NaF in 0.100 L.

K$_{}a$ = 3.5E-4 therefore pK$_{}a$=3.47

Homework Equations

pH=pK$_{}a$+log$_{}10$[base]/[acid]
(Henderson-Hasselbach equation)
and of course
-log$_{}10$(H$^{}+$)=pH

The Attempt at a Solution

I've tried an ICE table but my best guess for a solution was
0.01 M H+ (from HNO$_{}3$) react with the F$^{}-$ base ions to form HF.
.55 M( fluoride ions)-.01 M( hydronium)=.54 (final fluoride ions)
0.24 M (HF) + 0.01 M (HF produced from above reaction) = 0.25 M HF
So,
pH=3.47+log$_{}10$(.54/.25)
pH=3.80

The answer that is supposed to be correct is 3.79... is this just a sig figs error or did I mess up along the line because I needed exactly 3.79 to be the pH

Sorry if I did anything wrong, I'm new to this forum! (:

 

[Borek]

Don't use LaTeX just for subscripts - it doesn't work. It is better to either enter the expression entirely as LaTeX - so pK_a will look like $pK_a$ or to format it using [noparse][/noparse] and/or [noparse][/noparse] tags - so [noparse]HNO₃[/noparse] will become HNO₃.

Your approach is correct, just 3.47 is not correct enough. Try to calculate

$$-log(3.5\times 10^{-4})+log(\frac {0.54}{0.25})$$

in one step, or at least without rounding down intermediate values.

Note, that 3.79 doesn't make much sense - neither concentrations nor Ka are given with accuracy high enough to justify giving answer with three significant digits. We often report calculation of pH with two decimal digits, but we rarely should.