Question

For a titration experiment: I am using 1.025 M of NaOH(aq) as my titrant.

My titration solution is 4.0 g of potassium hydrogen tartrate, with 300 mL of distilled water. After 4 trials, my average NaOH used was 1.680 mL. I am having problems figuring out how to calculate my findings. Each trial started with 50 mL of potassium hydrogen tartrate solution, and 2 drops of phenophthalein. Then I added 1.680 mL of NaOH (on average) to get titration to endpoint. Temp for the solution was 23.2˚C.

1- Calculate the total volume and moles of NaOH required to reach the endpoint for each trial. Calculate the molar solubility of potassium hydrogen tartrate (in mol/L) for each trial and include that in the Excel table and calculate Ksp for potassium hydrogen tartrate.

Please show me steps on how to calculate the answers for future problems (as I have to use this on a few other experiments)

Thank you in advance.

Answer 1

The moles of NaOH are calculated:

n NaOH = M * V = 1,025 M * 0.00168 L = 1.72E-3 mol

The molar solubility is calculated:

S = n NaOH / V KHP = 1.72E-3 mol / 0.3 L = 5.73E-3 M

Ksp is calculated:

Ksp = [H +] * [KP-] = 5.73E-3 * 5.73E-3 = 3.29E-7

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