please help I need it as soon as possible trial 2 was not done the over all average for part one was 26.66 and part 2 average was 5.82 from the whole class

Using 0.405 M NaOH for titration 25.00 mL saturated solution of KHT in 0.200 M NaCl 30 Trial1. 25 mL saturated solution NaOH used mL NaOH used ml Trial2. 25 mL saturated solution 26.66 Average NaOH = ml Show the Calculations: 1.1 Write an equation of the molar solubility of KHT (=[HT-) } [+] [HT] = Show the Calculation: [HT] = 1.2 Write an equation for the Ksp of KHT Ksp = [K+] [HT-] Show the Calculation: Ksp = 1.3 Calculate the percentage error using the known literature value at 18 C of Ksp =3.8 x10-4 Part II Solubilty of KHT in 25.00 ml NaCl with added 0.100 KCI Using 0.405 M NaOH for titration 4.0 Trial1. 25 mL saturated solution NaOH used mL NaOH used mL Trial2. 25 mL saturated solution 5.82 Average NaOH = mL 2.1 Write an equation of the molar solubility of KHT {=[HT-] } [HT-) = Show the Calculation: (HT-) = 2.2 Write an equation of Ksp of KHT Ksp Show the Calculation: Ksp = 2.3 What effect did adding KCl to the solutions have on the solubility of KHT and on the solubility product constant of KHT