(B). Determine the compound (or the element) that is being oidized and reduced i

Question

(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?

Explanation / Answer

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

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