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if the H+ in acid mine drainage is 6.3 x 10^-5 mol/L, the pH of the drainage water is ___.

It would require ___ gal of 0.1 M
Ca(OH)2 (2 is a subscrip) to neutralize the proton acidity in 1000 gal of such drainage water.

2 answers

pH = -log(H^+) = -log 6.3E-5 = ?

Ca(OH)2 + 2H^+ ==> 2H2O + Ca^2+
Convert 1000 gallons to L.
Then L x 6.3E-5 mol/L = mol H^+.
Mols Ca(OH)2 = 1/2 that (from the coefficients in the equation).
M Ca(OH)2 = mols Ca(OH)2/L Ca(OH)2. Yo know M and mols; solve for L and convert that to gallons Ca(OH)2.
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