Question

1 and 2 please

-755" 14-ANALYSIS OF SODA ASH EXERCISES 1. 28.92 mL of NaOH were required to titrate 0.5982 g of KHP to the phenolphtha- lein end point. 22.41 mL of the same NaOH solution were needed to titrate 25.00 mL of the HCl solution to the phenolphthalein end point. If 34.62 mL of HCl solution were needed to titrate 0.5879 g of the soda ash sample, then what is the percent Na, co, in this soda ash sample? 2. If 24.50 mL of NaOH are needed to titrate 0.4451 g of dry KHP, what is the molarity of the NaOH solution? 3. If the weighing of the KHP in question 2 were incorrect (0.0100 g too high), then calculate the true molarity of the NaOH solution, the percent error, and the error in ppt of the molarity in question 2 compared to the actual (true) molarity of the NaOH solution. Note: % error = experimental value - true value true value

Answer 1

11) mas KHP = 0.5962 g. Molar mass = 204 g/mol moles KHP = mass a 0. 5982 g molosmars 204 & more

1 = 2.932x10 3 mol [OTCOOH = 2.182x10 3 mol OoNa "Cook T NaOH & Lolar +H₂O Inaoh = 1kHP = 2.182x10 MNAOH = . = 2.182x10 mol 24.5x103 L "KHP- Nach = molarity X volume MnaoH = 2.932xlo mo! 28.92 x 103 L = 0.1014m NaOH = 0.08905M NaOH + HCl + NaCl + H₂O Mix = M₂ V₂ NaOH Hel O. 10148 22.41 m2 = M*25 ml IMACI = 0.0909M Thai = 0.090 9 mol x 84.62mX > 3.147 Anmol = 3.147x10 3 mol Nazaz + 2H1 2 Nacı + 110 & CO Nazloz = nHCl = 3.147 xlomol = 1.573x10 3 mol mass = moles x molar mass = 1:573X163 mol X 1069 mol 5 16689 %. Na CO2 = 0,1668 Xloo 0.5879 = (28.37.1 | Na coz is present in soda ish sample, "Maoy = 24.50ML m = 0.445 n=0.44519 204 amor!