Question

The Na,Co, solution described above was used to analyse a hydrochloric acid solution of unknown concentration. It was found that 24.30 mL of the Na,Co, solution was required to neutralise a 20.00 mL sample of the hydrochloric acid solution. 4. Using your answer to part 3 above, calculate the amount of Na,Co, present in 24.30 mL of the Na,Co, solution. (2 marks) 5. Using your answer to part 2 above, deduce the amount of HCl that was present in the 20.00 mL sample of the HCl solution. (2 marks) 6. Using your answer to part 5 above, calculate the amount concentration of the HCl solution. (2 marks) 7. Assuming that 1 mol of CO, Occupies 24.1 L at room temperature and pressure (i.e. that the molar volume is 24.1 L mol), calculate the volume of CO2 formed in this reaction. (2 marks) 8. Calculate the pH of the HCl solution. (4 marks) 9. Strictly speaking, it is not correct to say that the solutions described above contain HCl or Na2CO3. Explain why, specifying what species are actually present in these solutions. (4 marks)

Answer 1

Answere feid : It is defined as a substance capable of donating a proton en a reaction. Conjugate base: It is the species foremed when an acid looses its proton. salt. It is the compound fromed from the reaction of an acid with a base .. Now, if we look into the given reaction, we find out that hee acts as the pecoton donors so it is the acid. After Hee looses its preoton, ce is left sit is the conjugate base. Nace is the sale formed due to neutrcalization Na2CO3 reaction between Hel Cacid) and Aatico, (base) - 2 H cecast Na2CO3 caq? 2 Nace caq) + H₂Oce) & Coacg) The above equation has been balanced where 2 equivalent of Hee reacts with & equivalent of Na₂CO3 to produce 2 equivalent Nace , lequi- -valent of H₂O and co2 each. Amount of Na, coz dissolved = 5.009 Amount of water added = 250 mL. Molecular weight of Na2CO3 = 106 g/mol

106 g of Na₂CO3 en 1000ml H₂O = 1CM) Ig of Na₂CO3 in 1000ml H₂O = 1 CM) x 1g 1064 i. 5.00g of Na₂CO3 in 250ml H₂O = 1 CM) x 59 x 250ml 1069 2000m 20.012(M) wy Fore neutrealization, 24.30mL of 0.012 CM) Na, co, was required. 250 mL of 0.012CM) NA, Coz = 5.00g .. 24.30 m L 040.012 CM) Na, C03 - 5.00g x 24-30 me 250ml. = 0.486g ~0.49g

5. From part 2 Hel: Na coz = 2:1. . Amount of Hel = 2x Amount of Na₂CO3 - 2 x 0.499 0.989 6. Molecular weight of Hel = 36.5 g/mol 1000mL $CM) HCl = 36.5g. .: 20mL 0.98g Hel = 1 CM) X 18g x20mL & 36.5g 100omL = 0.00053CM) = 5.3x10-4CM) 7. From the reaction, it is clear that 1 mol of Na, CO₂ reacts to produce 1 mol of co₂Cg) Thus, 24.1 L of co2 is focimed in this reaction. 8. The formula fore calculating plt is a pH = -log[MT] : [H+] - concentration of Htions => pH = -log(5.3810-4) » pH = 3.28 9. Both Hee and Na, coz are strong electrolytes. So, when they are becought into solution, they dissociate completely to the constituting ions and then the ions get solvated by the water molecules Helcaq) - Htcaq) + ce-caq Na, co , cag) 2 Na + (aq) + co 2-caq)