Post a New Question


posted by on .

A home swimming pool is disinfected by the daily addition of 0.50 gal (1 gal = 3.785 L) of a "chlorine" solution–NaOCl in NaOH(aq). To maintain the proper acidity in the pool, the NaOH(aq) in the "chlorine" solution must be neutralized. By experiment it is found that about 220 mL of an HCl(aq) solution that is 31.4% HCl by mass (d = 1.16 g/mL) is required to neutralize 0.50 gal of the "chlorine" solution. What is the [OH–] of the "chlorine" solution?

  • chemistry - ,

    g HCl added = 220 mL x 1.16 g/mL x 0.314 = ?
    mols HCl = ?/36.5 = x.
    M = mols/L = [xmols/(1/2)*3.785] = (OH^-)
    Check my work carefully.

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question