Michael S. answered • 08/05/26
Chemistry B.S. from Indiana University, A's in College Physics I & II
Ksp = 6.7 × 10−3
The dissolution equilibrium
KHC4H4O6(s) ⇌ K+(aq) + HC4H4O6−(aq)
Ksp = [K+][HC4H4O6−] — the solid does not appear.
Step 1 — what the titration actually measures
The bitartrate ion is the species with the acidic proton, and it is monoprotic, so it reacts 1:1 with hydroxide:
HC4H4O6− + OH− → C4H4O62− + H2O
mol NaOH = (0.0082 L)(0.055 M) = 4.51 × 10−4 mol
So the 5.5 mL sample contained 4.51 × 10−4 mol of bitartrate.
Step 2 — concentration in the saturated solution
Divide by the volume of the sample, 5.5 mL — not by the titrant volume, and not by the sum:
[HC4H4O6−] = 4.51 × 10−4 mol / 0.0055 L = 0.082 M
Step 3 — get [K+] free
The salt dissolves 1:1, so every bitartrate ion in solution arrived with a potassium ion:
[K+] = [HC4H4O6−] = 0.082 M
This is the step that makes the whole experiment work — you only titrate one ion, but the stoichiometry hands you the other for nothing.
Step 4
Ksp = (0.082)(0.082) = 6.7 × 10−3
Assumptions worth stating in your lab report
1. The solution was saturated and filtered free of solid. Undissolved crystals carried into the aliquot would titrate as well and inflate your result.
2. It was made in pure water, with no common ion. If your procedure used KCl solution instead, [K+] is not equal to the titrated concentration — you would add the KCl contribution, and Ksp comes out different.
3. Temperature was recorded. Ksp is temperature-dependent, and bitartrate is noticeably more soluble warm — which is exactly why cream of tartar crystallises out of wine as it cools in the cellar.
Reasonableness: 6.7 × 10−3 is a large Ksp compared with something like AgCl (1.8 × 10−10), which fits — potassium bitartrate is only slightly insoluble, not dramatically so. Published values sit near 3 × 10−4 at 25 °C, so a result several times higher usually points to an unsaturated-versus-warm solution or carryover of solid, both worth mentioning in your discussion.
