Answer to Question #176473 in General Chemistry for Lawrent

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 $=\frac{6799×100}{1000}= 679.9 Kg$