# Adding a strong acid to buffer solution

1. Mar 14, 2012

### kevvyroo

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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( flouride ions)-.01 M( hydronium)=.54 (final flouride 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! (:

2. Mar 14, 2012

### Staff: Mentor

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.