Please help! It\'s homework for a geology assignment but it\'s all chemistry bas

Question

Please help! It's homework for a geology assignment but it's all chemistry based and I haven't gotten through all of my chemistry classes yet so I'm struggling. Thank you!

Dissociation of Weak Acids Background Information Modality (m): the number of moles of a solute per kg of H20 Formality (F): the number of moles of a solute per kg of solution Molarity (M): number of moles of solute per liter of solution Normality (N): number of equivalent weights of solute per liter of solution When a chemical reaction between ionic species and molecules in a solution ha achieved equilibrium, the Law of Mass Action applies with the following cooventions l. The activities of ions and molecules must be expressed in terms of moles, but may be referred to in as a unit weight of solvent (molality), a unit weight of solution (formality), or a unit of volume of a solution (molarity) ll. The activities of water or pure solids are equal to one. lll. The concentration of a gas is expressed as the partial pressures in atmospheres. TV. Reactions are assumed to take place at STP (25°C; 1 atm) unless otherwise indicated Problems Write the balanced chemical equation for the non-metal oxide below as it interacts with water l) CO2 2) SO2 3) SiO2 4) If the concentration of an ion in solution is 5.0 x 10 mol/Lat 25°C, what is the concentration in the same solution at 45oC? Use the data in Table 9. l (Faure, 1998) 5) Calculate the pH of hydrofluoric acid containing 0.1 mol of HF per liter of solution. Use table 9.3 (Faure, 1998) Hydrochloric Acld 6) Calculate the normality of hydrochloric acid Acide Chlorohydrique (HCl) for the following label: Trace Metal Grade Assume the following: volume 2.5L; net wt 2.72kg molecular wt 36.46 g/mol specific gravity 1.18 kg/m T 25°C, P 1 atm

Explanation / Answer

1) CO2 + H2O -----> H2CO3 (Carbonic acid)

2) SO2 + H2O -----> H2SO3 (Sulfurous acid)

3) SiO2 + H2O -----> H2SiO3 (Silicic acid)

5) [HF] = 0.1 mol/L or 0.1 M

pKa of HF (from table 9.3) = 3.2

pKa = -log Ka

Ka = 10-pKa = 10-3.2 = 6.3 X 10-4

Ka = [H+] X [A-] / [HA]

6.3 X 10-4 = [H+] X [H+] / 0.1 (Assume [H+] = [A-])

[H+]2 = 6.3 X 10-4 / 0.1 = 0.0063

[H+] = 0.0793 M

pH = - log [H+] = - log 0.0793 = 1.1

pH of HF= 1.1

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---