Lab 7 CHEM 102 Worksheet PDF

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chemistry lab 102...

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Noemi Maldonado Jadeana M. Lab 7 – Determination of the Solubility Product Constant of an Organic Salt Goal: The goal is to determine the Ksp of KHT.

Table 1: The figure above shows two trials pf titration taking place between KHT ans NaOH.

Temperature of Filtered KHT sol (℃) Concentration of NaOH sol (mol/L) KHT solution Final Reading (mL) Initial Reading (mL) Total Volume KHT used (mL) NaOH solution Final Reading (mL) Initial Reading (mL) Total Volume NaOH used (mL)

Determination 1

2

19.7

19.7

0.04375

0.04375

35.00 0.00 35.00

35.00 0.00 35.00

28.35 0.00 28.35

27.55 0.00 27.55

Analysis List equations Used for Analysis KHTK+ (aq) +HT-(aq) Eq.2 Ksp=[K+][HT-] Eq.3 # moles of NaOH= (concentration of NaOH mol) x (total volume NaOH used(L) Eq.4 [HT-]= # moles of HT)/(volume of KHT solution, L) Eq.5 Number of moles of HT- = number of moles of HT- = number of moles of NaOH Eq.6 Total volume used = final volume-initial volume 1st trial: # moles NaOH=(0.04375 mol L-1) x(28.35 x10-3L)= 1.240 x10-3 mol 2nd trial: #moles NaOH=(0.04375 mol L-1) x (27.55 x10-3L)= 1.205 x10-3 mol Table 2: TITLE

Number of moles NaOH used (mol) Number of moles HT - used (mol) [HT] in KHT solution (mol/L) [K+] in KHT solution (mol/L) 1

Determination 1 1.240 x 10-3 mol 0.035 mol/L 0.035 mol/L 0.035 mol/L

2 1.205 x10-3 mol 0.034 mol/L 0.034 mol/L 0.034 mol/L

Ksp Value Average Ksp (± stand. dev.) Ksp Literature Value Percentage Error (%)

2

1.225 x 10-3 1.191 x10-3 1.16  10-3 2.67%

1.156 x10-3

Discussion: One paragraph with at least 2 references.

Conclusion: One sentence. The average Ksp determined is 1.191 x10-3, and the percentage error determined was 2.67%.

References:

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