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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: myelver10 on December 17, 2004, 04:08:38 AM

Title: strong base added to strong acid
Post by: myelver10 on December 17, 2004, 04:08:38 AM

what is the pH of the solution that results from adding 15 mL of 0.50 M NaOH to 25 mL of 0.50 M HF?  Ka for HF = 7.2 * 10^-4      

Title: Re:strong base added to strong acid
Post by: AWK on December 17, 2004, 07:04:38 AM

Hint - this is a buffer solution

Warning - HF is a weak acid

Title: Re:strong base added to strong acid
Post by: pizza1512 on February 03, 2005, 09:02:54 AM

Quote from: AWK on December 17, 2004, 07:04:38 AM

Hint - this is a buffer solution

What that?.



 ???

Title: Re:strong base added to strong acid
Post by: jdurg on February 03, 2005, 05:10:31 PM

It's a buffer solution because there's not enough NaOH to fully neutralize the HF in the mixture.  As a result, you wind up with a mixture of HF and NaF which is a weak acid and a salt of its conjugate base.  That equals a buffer.

Title: Re:strong base added to strong acid
Post by: Kong on February 04, 2005, 05:51:15 PM

very good.
A common misconception is that chemistry students think that once an acid/base is neutralized....it is neutral.   Wrong.  Only strong acids and bases are actually neutral.  Meaning the conjugates of strong acid/bases are neutral.   HF is a weak acid so its conjugate F- is a weak base.  
AS

Title: Re:strong base added to strong acid
Post by: jdurg on February 07, 2005, 07:48:53 AM

Very true as well.  Though in this case, if there was enough NaOH to fully react with the HF I'd feel safe in calling the overall mixture 'neutral'.  Mostly because I don't think the F- ion will tend to pull an H+ off of water and reform HF all too frequently.  If it did, then nearly all fluoride solutions would attack glass.