 # EXPERIMENT #2

NaOH Standardization and Determination of KHP Purity

Preparation of NaOH/Identification of Ukw

Experiment Summary:  The ultimate objective is to discover the purity of an unknown solid acid formed by mixing KHC8H4O4 with a non-reactive salt.  A sodium hydroxide solution of 0.1206M was ready from the previous lab, and this was used to titrate samples of the unknown mixture.

# Preparation of NaOH solution

A standardized solution was left over from the previous lab (lab 10).  I used this 0.106 M NaOH solution to do the titration on the acid of unknown strength (the KHP).

# Purity Determination of Unknown (KHC8H4O4)

The unknown salt compound provided was divided into four samples.  These samples (detailed in Table A) were titrated with the Secondary Standard, NaOH.  Although three of the titrations did not result in viable results, the fourth reached the endpoint cleanly.  The resulting volumes of NaOH needed were used to computer the number of moles of KHC8H4O4 present in each of the parts.  Then, the moles were converted to grams, and the mass percentage was calculated (mass-HC8H4O4/mass-sample *100%).

# Sample Calculations

## Finding the Molarity of the NaOH Secondary Standard:

The molarity of the sodium hydroxide solution was known to be 0.1206M from the previous lab’s results.

## Finding the concentration of the unknown KHC8H4O4 mixture: TRIAL “N”

I converted the volume of NaOH to Liters (0.01981 L), and then used the molarity (0.1206M) to convert it to moles (0.001201 moles).  Because the equation of neutralization uses one mole of each reactant (NaOH and KHC8H4O4), the number of moles of KHC8H4O4 in the sample is also 0.001201oles.   We can then find the percentage of the sample that consists of KHC8H4O4 by finding the mass of that number of moles.  We do this by multiplying the number of moles of KHC8H4O4 by the molar mass of KHC8H4O4, 204.23 g/mole.  This tells us that there were 0.2453 grams of KHC8H4O4 in the sample.  The mass percentage is the mass of the part divided by the mass of the whole times 100%; this is 0.2453g/0.7346g * 100%=33.39% KHC8H4O4 by mass.

# Table 2: The Unknown Concentration

 Trial Volume of NaOH mL Mass Ukw. Sample grams Mol in sample - KHC8H4O4 Mass (g) in sample -KHC8H4O4 Mass % K. A 10.63 0.7466 g (overtitrated) B 9.68 0.7035 g (found a bubble in the buret, overtitrated) C 9.85 0.7053 g (overtitrated) N 9.96 0.7346 g 0.001201 mol 0.2453 33.39%

The % mass for the unknown salt was calculated to be 33.39% KHP by mass.

# Analysis of Results

Many of these samples were overtitrated.  This was caused by a solution of NaOH which was too strong for the titrations to proceed cleanly.  To solve this problem, the NaOH standard could be diluted using a pipet and a volumetric flask so that the molarity can be determined without titration.  Alternatively, after the first titration was complete, the chemist (me) could have mixed up a custom solution of NaOH designed to titrate the unknown solution.  When the volumes for the titrations are so small, the error in reading becomes more significant in the results.  Overtitration by even one drop yields a much more significant error than would take place with a larger volume of titrant.  The final titration went really well, though (the endpoint solution was a beautiful shade of pink), and I am confident that the result (33.39% KHP by mass) is very close to the true value.