Question

Sulphuric acid reacts with sodium hydroxide as follows.H2SO4 + 2NaOH ⇒ Na2SO4 + 2H2Ohen 1L of 0.1 M sulphuric acid solution is allowed to react with 1 L of 0.1 M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is

• A. (A) 0.1 mol L-1
• B. (B) 7.10 g
• C. (C) 0.025 mol L-1
• D. (D) 3.55 g

Answer 1

Answer: (B), (C)

(B) 7.10 g

Answer 2

Explanation:
1L of 0.1MH2SO4 contains 0.1molH2SO41 L of 0.1M. NaOH contains =0.1molNaOH According to the given equation, 1 mol of H2SO4 reacts with 2mol of NaOH.Therefore, 0.1 mol of NaOH will react with (0.12)=0.05mol of H2SO4 and 0.05 mole H2SO4 will remain unreacted. Thus, NaOis the limiting reagent. ∴0.1 mole of NaOH will produce 0.05 mole of Na2SO4Molar mass of Na2SO4=2×23+1×32+4×16=142gMass of Na2SO4 formed =142×0.05=7.10gVolume of solution after mixing =2L H2SO4 left unreacted =0.05molMolarity of Na2SO4 formed =0.052=0.025molL−1