Assume that the molarity of the NaOH solution used to titrate the saturated solu

Question

Assume that the molarity of the NaOH solution used to titrate the saturated solution was actually higher than reported. Would the calculated KHT solubility be incorrectly high or low? Briefly explain. Assume that the filter paper had a small hole in it, allowing some solid KHT to get into the saturated solution filtrate. If some of the solid were included in the portion of solution titrated with NaOH solution, what effect would the solid have on the calculated KHT solubility? Briefly explain. An experiment was performed to determine the K_p of Ca(OH)_2. A solution containing 2.00 times 10^-2M Ca(NO_3)_2 was saturated with Ca(OH)_2 and filtered. Two samples of the saturated solution were titrated with 5.021 times 10^-2 M HCI, in order to determine the OH^- ion concentration in the saturated solution. In doing this problem, note that 1 mol of HCL reacts with 1 mol of OH^- ions. This relationship is the basis for the determination of the OH^- ion concentration: 1 mol Ca_2+ ions are produced for every 2 mols of OH ions produced. The data resulting from this experiment are shown in Table Calculate the following: (a) number of moles of OH^- ions titrated, mol (b) |OH^-| (c) |Ca^2+|_initial d) |Ca^2+|_from dissolved Ca(OH)_2 (e) |Ca^2+| (f) K_sp (g) average K_sp

Explanation / Answer

5. Let's consider KHT(Potassium Hydrogen Tartarate)(s)=K++HT-(aq)

If solubility of KHT(s) is S then we can write

S=[molarity of NaOH used for titration}*[1 mol HT-]/[1 mol NaOH]

As per the given question [molarity of NaOH used for titration] is higher than reported. So eventually, calculated KHT solubility is incorrectly high.

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