5 Titration of Drain Cleaner Given concentration of standard

5. Titration of Drain Cleaner Given concentration of standard HO (M) Mass of deaner () g 6S 62 G so pt at first equivalence point Number of drops of NC at first equivalence point Velume of HOI at fist equivalence peint (cakulate) (ml ph at second equiralkence point Number of drops of HCI at second equivalence point Volume of HO at second equivalence point (akulate) (ml) 20 Write the balanced reaction equations for the titration of (a) NaOH with HCl and (b) NaOCI Calculate the moles of NaOH, the mass of NaOH, and the mass percent of NaOH in the cleaner Calculate the moles of NaOCI, the mass of NaOCl, and the mass percent of NaOCI in the cleaner. Think carefully about what volume to use; the volume is not directly listed in the data table. 50 CHM 230 Lab5

Solution

1) For NaOH

You required 6.504 ml of HCl with a concentration of 0.0963 M

moles = Molarity * volume

moles of HCl = 0.0963 * 0.006504 = 0.0006263 moles of HCl

you require 1 mole of NaOH for every 1 mole of HCl so you have

0.0006263 moles of NaOH, molar mass of NaOH is 40 g/mol

mass = moles * molar mass = 40 * 0.0006263 = 0.025 grams of NaOH

For NaOCl, procedure is similar

you required 12.516 ml of HCl with a concentration of 0.0963 M

moles = Molarity * volume = 0.012516 * 0.0963 = 0.0012053 moles of HCl

1 mole of NaOCl requires 2 moles of HCl so

0.0012053 / 2 = 0.0006026 moles of NaOCl available

molar mass of NaOCl is 74.44 g/mol

mass of NaOCl = moles * molar mass = 74.44 * 0.0006026 = 0.04485 grams of NaOCl

Total mass NaOH + NaOCl = 0.04485 + 0.025 = 0.06985 grams

mass percent NaOH = 0.025 *100 / 0.06985 = 35.8 %

mass percent NaOCl = 100 - 35.8 = 64.2%

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