Question

This experiment has involved the study of the variation of solubility of potassium hydrogen tartrate with ionic strength and also determine its thermodynamic solubility product. The background and procedure are given below as a guide in answering the questions:

In the second experiment the electrolyte contains an ion in common with the saturating salt. The sparingly soluble salt is po

Figure 1 (above): Background Theory

1 Filter each of the KCl and (KCI NaCI) solutions through a dry No.40 filter paper rejecting the first 10 cm* 2. Pipette 10 c

Figure 2 (above); Procedure for handling NaCl + KCl and KCl only

KCl Concentration (in M) 0.01 M NaOH Added (in mL)
0 46.2
0.025 34.1
0.05 19.3
0.10 15.5
0.20 8.3
0.30 5.5
NaCl + KCl Concentration (in M) 0.01 M NaOH Added (in mL)
0 35.1
0.025 24.2
0.05 20.4
0.10 14.1
0.20 8.2
0.30 6.2

Table 1: Obtained results from titration of KCl and NaCl+KCl

QUESTIONS:
A.
From the titrations determine the molarity of tartrate in solutions (c2) and hence the mean ionic molarity where c1 is the molarity of potassium chloride in solution.

B. Plot the square root of I (taking into account the potassium hydrogen tartrate in solution) against log c\pm (ordinate). Extrapolate the line, determine Ks and hence \gamma \pm at each ionic strength (Equation (8)).

C. For both series, plot the square root of I against log \gamma \pm. Explain the results.

In the second experiment the electrolyte contains an ion in common with the saturating salt. The sparingly soluble salt is potassium hydrogen tartrate and the common ion is K The thermodynamic solubility product of uni-univalent potassium hydrogen tartrate KHC4H406 is: =a,a 0.5 Thus The graph oNI against log c, is linear, the intercept at 1-0, γ -1, is log VKs and hence γ+ can be calculated at any ionic strength.
1 Filter each of the KCl and (KCI NaCI) solutions through a dry No.40 filter paper rejecting the first 10 cm* 2. Pipette 10 cm3 aliquots of each solution and dilute with 20 cm3 of deionised water. Titrate in duplicate with the standard 0.01M NaOH solution using phenolphthalein as indicator.



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Answer #1

Organic compounds are often identified using spectroscopy. The process of testing compounds using spectroscopy is fairly simple (the compounds are placed into the machine and the read-out is printed). The difficult part comes in learning how to read the print-out and determining what it is telling you.

In infrared (IR) and nuclear magnetic resonance (NMR) spectroscopy, we identify the compound based on specific groups. Certain ranges of frequencies for each type of spectroscopy indicate different groups

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