Question

Sodium carbonate, Na2COs, reacts with hydrochloric acid, HCl, to produce sodium chloride, carbon dioxide and water. Refer to slide 7.18 for a summary of formulae relevant to the calculations below. 2HCI(aq) Na,cO3(aq) NaCl(aq) H2O(I) CO2(g) 1. Use this reaction to explain what is meant by the terms "acid", "conjugate base" and "salt". (6 marks) 2 Balance the equation for this reaction. (2 marks) A solution was prepared by dissolving 5.00 g of Na,CO3 in water and adding water to give 250 mL of solution. Calculate the amount concentration (molarity) of that solution. 3. (6 marks) The NA2CO3 solution described above was used to analyse a hydrochloric acid solution of unknown concentration. It was found that 24.30 mL of the Na,COs solution was required to neutralise a 20.00 mL sample of the hydrochloric acid solution.

Answer 1

& Hol HCl: acid, Sat) Nacii Salt vaqcoz i Nag coz consugate base (& Nait coz & balance the equation I HO + Nay Co] 2 Nad of Heos (3) Molarity (ty = weight of Nascoz PL! X 1000 nolecular weight off NaCO3 Volunelmy != 5918 1000 Hi 70% * P50 ric 20.188677 rolarity of Naqcoz solution 2 0.1886M

balanced equation is NaQ Cog + 2 Hey we know that t 2 raut H₂o 4 con he Miz Molarity of Hu=? Vi= Volurie of Hal= hohe mos number of males of HCl = $ Me = Molanity of Nag Coz= 0.1886 M V = volume of Naqcoz = 240 30 rz nq = Moles of Narco3 = 1 . MA E Mevan, nevi 0.1886 x 24.30% 2 i 30.4582 . Molarity of Ha (14)=0.45897 E