Concentration calculation

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

show the calculation for [HT-] in your saturated solution for part 2.

titration- 14.30ml 0.04M sodium hydroxide
i used. 2.005g KHT and 8.5g NaNO3

i diluted the NaNO3 to 100ml and added 2g KHT. filtered it then titrated with sodium hydroxide.

2. Relevant equations

KHT---- K+ + HT-

3. The attempt at a solution

i figure i have to do a I.C.E chart to find this out and since KHT is a solid it has no effect.
and i do know that [K+]=[HT-]
it could be but i dont believe it would be that simple, i saw other people in the class and they used an ice chart i think.


Question is ambiguous then.

Titration was to determine solubility of tartrate. Once you get total concentration you need to account for two processes - hydrolysis and dissociation. Simple ICE table won't work here, and in general this is not an easy problem.

However, I think it is possible to solve the question using a trick. Please see caculation of pH of amphiprotic salt. Equation 12.6 holds always. If you will find values for Ka1 and Ka2, and if you will find information about pH of saturated KHT solution, you will be able to find exact concentration of HA-. It happens that all three numbers are well known, especially pH of the saturated solution was thoroughly researched.

Problem is, with this approach you don't use results of your experiment.
they next part of the question which i didnt post says to find the Ksp(solubility product) of KHT in you saturated solution.

so maybe i just use dilution to find [HT-] and then plug this into a ICE chart to find the Ksp.


You have diluted something? So far you only said about titrated, saturated solution.

And I am not sure I know how do you want to use ICE table for finding Ksp. Assuming there were no hydrolysis and dissociation, you have just [HT-]=[K+].

Advanced rant:

I have checked - unfortunately, you can't ignore hydrolysis and dissociation, final concentration of HT- is substantially lower than the one determined by titration.

This is often problem with simple labs, they ignore part of the chemistry involved and you never know if you are expected to take everything into account, or not.

i figured it out, thank you.

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