B Determine the compound or the element that is being oidize

(B). Determine the compound (or the element) that is being oidized and reduced in each reaction and write down the oxidation numbers for compounds being oxidized and reduced Oxidized Reduced i. Define the following terms Strong electrolyte Weak electrolyte Non electrolyte 13. A0250 M HPOs is required to neutralize a solution prepared by dissolving 17.5 g of NaOH in 350 mL of water i. write the balanced equation for the reaction of HjPwith NaOH. Calculate the number of moles of NaOH Calculate the number of moles of Hea iv. What volume (L) of 0.250 MHPOuis required to neutralize a solution prepared by dissolving 17.5 gof NaOH in 350 ml of water?

Solution

Ans. #BI. Oxidized: Cu(s).   Cu(s) ------> Cu(NO₃)₂.

Cu(s) with oxidation number 0 is oxidized into Cu(NO₃)₂ where Cu has an oxidation number of +2. Note that increase in oxidation number is called oxidation.

# Reduced: AgNO₃. Ag in AgNO₃ has oxidation number of +1. It is being reduced to elemental Ag, in which Ag has oxidation number of 0.

#BII. Strong electrolyte: An ionic compound that undergoes complete (100%) dissociation in aqueous medium is called a strong electrolyte. Example- NaCl, HCl, etc.

# Weak electrolyte: An ionic compound that undergoes partial (much lower than 100%) dissociation in aqueous medium is called a weak electrolyte. Example- acetic acid, HF, etc.

# Non electrolyte: A compound (usually covalent) which does not undergoes dissociation in aqueous medium is called a no-electrolyte. Example- glucose, sucrose, etc.

#13. I. Balanced reaction:    H₃PO₄(aq) + 3 NaOH(aq) ----> Na₃PO₄(aq) + 3 H₂O(l)

#13.II. Mols of NaOH = Mass / Molar mass = 17.5 g / (40.0 g/ mol) = 0.4375 mol

#13.III. Moles of H₃PO₄ = Molarity x Volume of solution in liters

Molarity alone is NOT sufficient to calculate mole of H₃PO₄. That is, insufficient data for this step.

#13.IV. According to the stoichiometry of balanced reaction of neutralization, 1 mol H₃PO₄ neutralized 3 mol NaOH.

So,

            Required moles of H₃PO₄ = (1/3) x Moles of NaOH in sample

                                                            = (1/3) x 0.4375 mol

                                                            = 0.1458 mol

# Required volume of H₃PO₄ = Required moles of H₃PO₄ / Molarity

                                                = 0.1458 mol / 0.250 M                   ; [1 M = 1 mol/ L]

                                                = 0.1458 mol / (0.250 mol / L)

                                                = 0.5832 L