Note : The solubility of potassium hydrogen tartrate (KHT, 188.18g/mol) is 1.00g

Question

Note :

The solubility of potassium hydrogen tartrate (KHT, 188.18g/mol) is 1.00g/162mL at 25C and 1.00g/16mL at 100C.

KHT (s) à K+(aq) + HT-(aq)

(PL1) Using the solubility given in the lab, calculate the solubility, in M, of potassium hydrogen tartrate at 25C and at 100C.

(PL2) Write an expression for Ksp of potassium hydrogen tartrate.

(PL3) Calculate the theoretical solubility product, Ksp, of potassium hydrogen tartrate at 25C and at 100C.

(PL4) Write a balanced equation showing the titration of potassium hydrogen tartrate solution with NaOH.

Explanation / Answer

1)

we know that

solubility in M = solubility in g/L / molar mass

At 25 C

solubility = 1 g / 162 ml = 1g / 0.162 L

solubility = 6.17284 g / L

now

solubility in M = solubility in g/L / molar mass

solubility in M = 6.17284 / 188.18

solubility in M = 0.0328

At 100 C

solubility = 1 g / 16 ml = 1g / 0.016 L

solubility = 62.5 g / L

now

solubility in M = solubility in g/L / molar mass

solubility in M = 62.5 / 188.18

solubility in M = 0.332

2)

KHT (s) ---> K+ + HT-

Ksp = [K+] [HT-]

3)

let the molar solubility be s

KHT (s) --> K+ + HT-

now

[K+] = s

[HT-] = s

Ksp = [K+] [HT-]

Ksp = s x s

Ksp = s2

At 25 C

Ksp = (0.0328)^2 = 1.07584 x 10-3

At 100 C

Ksp = (0.332)^2 = 0.11

4)

the reaction is given by

KHT + NaOH ---> NaKT + H20

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