Solution to Chapter 2 Homeworks

Solution to Chapter 2 Homeworks - 1 Chapter 2 Solutions Ch...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 2 Solutions Ch 2.5,9,15, 22, 33, 36, 56, 57 2-5 Concentration of sodium bicarbonate Given: 45.000 g of sodium bicarbonate in 1.00 L of water. Solution: a. The resulting concentration of the solution is 45 g/L. Converting this to mg/L (45 g/L)(1000 mg/g) = 45,000 mg/L b. To find the molarity, the molecular weight of NaHCO 3 must be found Na = 22.99 x 1 = 22.99 H = 1.008 x 1 = 1.008 C = 12.01 x 1 = 12.01 3O = 16.00 x 3 = 48.00 Σ = 84.01 g/mol Then, the molarity (45 g/L)(1 mol/84 g) = 0.536 M c. The equivalent weight of NaHCO 3 is its GMW divided by the number of hydrogen ions transferred. In this case n = 1 because Na + is replaced by 1 H. Thus normality (N) is n*M, and in this case n = 1 (0.536 M)(1) = 0.536 M d. Change to units of mg/L as CaCO 3 , using Eqn 2-87 50 mg/L as CaCO 3 45,000 mg/L *(-------------------------) = 26,785 mg/L or 2.68 x 10 4 mg/L as CaCO 3 84 mg/L as NaCO 3 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2-9 Concentration of ferric phosphate Given: 2.4 g ferric phosphate added to 1.0 L water, initial phosphate = 1.0 mg/L. Solution:
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 08/27/2009 for the course ENVE 331 taught by Professor Vigil during the Winter '08 term at Cal Poly.

Page1 / 10

Solution to Chapter 2 Homeworks - 1 Chapter 2 Solutions Ch...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online