100mL of .01M NaCl and 2 grams of KHTar **Section 3: Solution of KHTar in 0.10 M KCl****0.0500 M NaOH Volumes:**- **Initial:** - Measurement 1: 46.72 mL - Measurement 2: 47.26 mL- **Final:** - Measurement 1: 41.98 mL - Measurement 2: 42.64 mL- **Used:** - Measurement 1: _________ - Measurement 2: _________- **Average Used:** _________ mL**Instructions:**Show your method to calculate the \([HTar^-]\), \([K^+]\), and the solubility of KHTar in the KCl solution. Remember to include \(K^+\) from 0.10 M KCl.- \([HTar^-] =\) ____________- \([K^+] =\) ____________- **KHTar Solubility =** ____________**Task:**Calculate the \(K_{sp}\) value for KHTar in this solution.- \(K_{sp} =\) ____________________

Given: Concentration of KCl = 0.10 M The dissociation reaction of KCl is shown below. KCl(s) →K+(aq) + Cl-(aq) According to this reaction, 1 mol of KCl gives →1 mol of K+ So, the concentration of K+ in KCl = 0.1 MThe volume used is calculated by the difference in final and initial volume. The volume used = V (initial) - V (final) The volume used in trial A = 46.72 mL - 41.98 mL = 4.74 mL The volume used in trail B = 47.26 mL - 42.64 mL = 4.62 mL The average volume = 4.74 mL + 4.62 mL2= 4.68 mL The given concentration of NaOH = 0.0500 M The average volume of NaOH = 4.68 mL So, the number of moles of NaOH is calculated as, moles of NaOH = Molarity × Volume in Litre = 0.0500 mol/L × 0.00468 L = 0.000234 mol = 0.234 mmol The dissociation reaction of [KHTar] is shown below. KHTar(s) →K+(aq) + HTar-(aq) The reaction of [HTar-] in presence of NaOH is, [HTar-](aq) + OH-(aq)→[Tar-2](aq) + H2O(l) According to this reaction, the number of moles of [HTar-] is equal to the number of moles of NaOH. Thus, the number of moles of [HTar-] = 0.234 mmol. The given mass of [KHTar] = 2 g Since, 1 g = 1 mL 2 g = 2 mL So, the volume of [KHTar] = 2 mL = 0.002 L The molarity of [HTar-] = moles of [HTar-]Volume in Litre =0.000234 mol0.002 L = 0.117 M The concentration of [K+] = K+ (KCl) + K+ (KHTar) = 0.1 M + 0.117 M = 0.217 M The solubility of [KHTar] = 0.117 M The Ksp of [KHTar] is calculated as, Ksp = [K+][HTar-] = 0.217 M ×0.117 M = 0.0254 M2