Calculate the solubility of KHTar in three solvent systems: pure water, 0.10 M K

Question

Calculate the solubility of KHTar in three solvent systems: pure water, 0.10 M KNO3, and 0.10 M NaNO3. The solubility of KHTar was titrated of the HTar- with standard NaOH solution. Molarity of NaOH = 0.10M

Volume needed to titrate pure water = 10.2 mL Volume needed to titrate 0.100 M NaNO3 = 14 mL Volume needed to titrate 0.100 M KNO3= 5.5 mL 50 mL of solution was used for each titration From this data calculate the solubility of KHTar and the Ksp.

DO NOT GIVE ME AN OVERALL EQUATION AND EXPLANATION THAT I CAN GOOGLE FROM. PLEASE CALCULATE IT STEP BY STEP

Explanation / Answer

Volume of titrant was 50 mL solution to start with

The KHTar dissolves in water by the equation:

KHTar(s) <---> K+(aq) + HTar-(aq)

You titrated some volume of the saturated solution with NaOH to calculate the concentration of HTar- in the solution,

Since we have used 10.2 mL of 0.1 M NaOH KHTar in solution is

0.0102 x 0.1 = 0.00102 moles

0.00149 moles in 50 + 10.2 = 60.2 mL is

0.00102moles/0.0602L = 0.0169 M

Since HTar- is 0.0169 M K+ is also 0.0169 M

Ksp = [K+][HTar-]

Ksp = 0.0169 x 0.0169 = 2.87 x 10-4

In 10 mL NaNO3

Since we have used 14 mL of 0.1 M NaOH KHTar in solution is

0.014 x 0.1 = 0.0014moles

0.0014 moles in 50 + 14 = 64 mL is

0.0014moles/0.064L = 0.02187M

Since HTar- is 0.02187 M K+ is also 0.02187M

Ksp = [K+][HTar-]

Ksp = 0.02187 x 0.02187 = 4.78 x 10-4

Since we have used 5.5 mL of 0.1 M NaOH KHTar in solution is

0.0055 x 0.1 = 0.00055 moles

0.00055moles in 50 + 5.5 = 55.5 mL is

0.00055moles/0.0555L = 0.0099M

Since HTar- is 0.0099 M K+ is also 0.0099 M

Ksp = [K+][HTar-]

Ksp = 0.0099x 0.0099 = 9.82 x 10-5

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