Question

Lakes that have been acidified by acid rain can be neutralized by the addition of limestone. How much limestone in kg is required to completely neutralize a 4.3billion litre lake with ph of 5.5

Answer 1

From the given pH, we calculate the concentration of H⁺:

   $[H^+] = 10^{-pH} = 10^{-5.5}$

We then use the volume to solve for the number of moles of H⁺:

   moles $H^+ = 10^{-5.5}M ×(4.3×10^9)L = 13598 moles$

From the balanced equation of the neutralization of hydrogen ion by limestone written as

   CaCO3(s) + 2H⁺(aq) → Ca2⁺(aq) + H2CO3(aq)

we use the mole ratio of limestone CaCO3 and H⁺ from their coefficients, which is 1 mole of CaCO3 is to react with 2 moles of H⁺, to compute for the mass of the limestone:

   

moles CaCO3 $= \frac{13598}{2}=6799g$

  

Since molar mass of CaCO3= 100

 mass CaCO3 $=\frac{6799×100}{1000}= 679.9 Kg$