Question

If 25.0cm³ of 0.1 MH₂S0₄ solution neutralized a solution containing 1.06g of sodium carbonate in 250cm³ of solution, calculate,

1. The molarity of sodium carbonate solution (Na=23, 0=16,C=12)  
2. Volume of sodium carbonate solution used.   

Answer 1

1. Moles of Na₂CO₃
   R.F.M=106
   =^1.06/₁₀₆
   0.01moles
   If 0.01moles→250cm³
           ?         →1000cm³
   1000 × 0.01 =0.04M
        250
2. H₂SO₄ + Na₂CO₃→Na₂SO_4(aq) + CO_2(g) + H₂O_(l)
   From the r.r of 1:1
   Moles of H₂SO₄ used=Moles of Na₂CO₃
   Moles of H₂SO₄ used 
        0.1Moles → 1000cm³
            ?        → 25cm³
         0.1×25 =0.0025moles
           1000
    If 0.04moles →1000cm³
     0.0025moles→  ?
   0.0025 × 1000
            0.04
       =62.5cm³