Find pH at Endpoint: Formic Acid Titration with NaOH Solution - 8.32 Answer

[coookiemonste]

Homework Statement

10 mL of a solution of formic acid HCOOH required 31.23 mL of .1034 M NaOH to titrate it. Find the pH at the endpoint. Answer is 8.32.

Homework Equations

pH=pKa + log[A-]/[HA]

The Attempt at a Solution

initial concentration of HCOOH is .3229 M from fact .1034 M(.03123L)=.003229 mol NaOH, .003229mol HCOOH/.01L =concentration of HCOOH.
pka= 3.75

 

[symbolipoint]

A different viewpoint can be more helpful, starting with the salt as a base (opposite of "acid").

Ka*Kb=Kw, reaction being H2O + NaA <--------> Na+ + OH- + HA
but you only care about H2O + A- <------> OH- + HA.

Kw/Kb = ([OH-](Facid + [OH-])/(Fsalt - [OH-]) = [OH-][HA]/[A-]

(please make your own adjustments for my bad typesetting, or lack of good typesetting).

From the "K" line, you really want to focus on: Kw/Kb = ([OH-](Facid + [OH-])/(Fsalt - [OH-])

Facid means the formal concentration of the weak acid, which is extremely small at the endpoint.
Fsalt means the formal concentration of the sodium formate, as if the titration solution were prepared only from this salt alone without use of acid or base.

 

[Borek]

At the equivalence point (which is what you can calculate, end point is when you have detected titration should stop) you have a solution of formate, so you just calculate pH of salt. That means - as symbolipoint suggested - treating solution as solution of a weak base HCOO^- and calculating [OH^-], then converting it to pH.

Much more here:

acid base titration equivalence point calculation

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