25 cc of an alkali solution is mixed with 8 cc of 0.75 N acid solution for complete neutralization it further
requires 15 cc of 0.8 N acid solution. Find the strength of the given alkali solution
Answers
Explanation:
The specific gravity of sodium carbonate is 1.25 g/mL.
Hence, the mass of 25 mL of solution of sodium carbonate is 1.25×25=31.25 g.
The mass of HCl used for neutralization is
1000
32.9
×109.5=3.60 g.
The number of moles of HCl in 3.60 g is
36.5
3.60
=0.0987.
2 moles of HCl will neutralize 1 mole of sodium carbonate.
The number of moles of sodium carbonate neutralized by 0.0987 moles of HCl are
2
0.987
=0.04935
Thus, 31.25 g of sodium carbonate contains 0.04935 moles.
Hence, 125 g of sodium carbonate will contain
31.25
125
×0.04935=0.1974 moles.
They will neutrlaize 0.1974 moles of sulphuric acid which corresponds to 2×0.1974=0.3948 g eq of sulphuric acid.
The volume of 0.84 N sulphuric acid required will be
0.84
0.3948
=0.470 L or 470 mL.
Answer:
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