Plz help me in this problem.i wud appreciate.?

Question

Lakes that have been acidified by acid rain can be neutralized by liming, the addition of limestone [CaCO3].

How much limestone (in kg ) is required to completely neutralize a 4.2 billion liter lake with a pH of 5.3?

Answer 1

-log10(x) = 5.3

log10(x) = -5.3

x = 10^(-5.3) = 5.01 X 10^(-6) moles of H+ per litre of solution.

Moles of H+

=   5.01 X 10^(-6) X 4.2 X 10^9

= 21049.86381 moles

= 21050 moles (approx)

Therefore moles limestone needed

= 21050/2

= 10525 moles

Now, mass of 1 mole of limestone

= 40 +12 + 48 = 100 grams

Therefore mass of limestone needed

= 10525 X 100 g

= 1052500 g

= 1052.5 kg

= 1.053 tonnes