PART $1$: Determination of the Molar Solubility and Ksp of KHT in Water Actual concentration of NaOH solution $($MNaOH$)$ KHT sample volume $($VHT $)$

Part $1$: Table of Data $($all values should have at least $3$ significant figures$)$

Run $1$

Run $2$

Run $3($optional$)$

initial NaOH burette volume final NaOH burette volume

NaOH titrant volume $($VNaOH$)$

$[$KHT$]=[$HT$]$

average molar solubility of KHT $[$K$+]$

Ksp

average Ksp of KHT

CALCULATIONS

$$ For $[$HT $]$ use titration Equation $11\mathrm{.}2$

$$ Molar solubility $=[$HT$]=$ moles of KHT dissolved per liter of solution.

$[$K$+]=[$HT$]$

$$ Use Equation $11\mathrm{.}1$ for Ksp

$151$

General Chemistry II Laboratory, Experiment $11$

Solubility Product of a Compound

PART $2$: Determination of the Molar Solubility and Ksp of KHT in $0\mathrm{.}05$ M KCl$.$

Preparation of $100$ mL of $0\mathrm{.}05$ M KCl Solution. Show Work Below.

Step $1$: Calculate the number of moles of KCl needed $($molarity$$volume$).$ Step $2$: Convert moles of KCl into grams of KCl using molar mass.

Mass of KCl that you calculated$/$weighed

Actual concentration of NaOH solution $($MNaOH$)$ KHT sample volume $($VHT $)$

Part $2$: Table of Data $($all values should have at least $3$ significant figures$)$

Run $1$

Run $2$

Run $3($optional$)$

initial NaOH burette volume final NaOH burette volume

NaOH titrant volume $($VNaOH$)[$HT$]$

average molar solubility of KHT $[$K$+]$

Ksp

average Ksp of KHT

Calculations are the same as for Part $1$ except that $[$K$+]=[$HT$]+0\mathrm{.}05$ M$.$