which is cheaper :40%HCl at the rate of 50 paise per kg or 80%sulphuric acid at the rate of 25 paise per kg to completely neutralize 7 kg of caustic potash?

Reaction of HCl with NaOH and H_{2}SO_{4}:

HCl + KOH → KCl + H_{2}O

36.5 56

H_{2}SO_{4 }+ 2KOH → K_{2}SO_{4} + 2H_{2}O

98 56 x 2

It can be seen that

56 g of KOH is neutralized by 36.5 g of HCl

Therefore,

7000 g of KOH will be neutralized by 36.5/56 x 7000 = 4562.5 g = 4.56 Kg HCl

So price of 4.56 Kg HCl = 4.56 x 50 paise = 228 paise

And

56 x 2 = 112 g of KOH is neutralized by 98 g of H_{2}SO_{4}

Therefore,

7000 g of KOH will be neutralized by 98/112 X 7000 = 6125 g = 6.12 Kg of H_{2}SO_{4}

So price of 6.1 Kg H_{2}SO_{4}= 6.1 x 25 paise = 152.5 paise

Thus, H_{2}SO_{4} is cheaper.

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