Question
1. The equation for the reaction of aqueous phosphoric (V) acid, H_(3)PO_(4), with aqueous sodium hydroxide, NaOH(aq) is shown below. H_(3)PO_(4)(aq)+3NaOH(aq)arrow Na_(3)PO_(4)(aq)+3H_(2)O(l) 25.0cm^3 of a 0.200moldm^-3H_(3)PO_(4)(aq) is titrated with 0.600moldm^-3NaOH(aq) Which statement is correct? A. The end point occurs when 25.00cm^3 of NaOH(aq) has been added. B.The end point occurs when 75.00cm^3 of NaOH(aq) has been added. After titration the final solution contains 0.0150 mol of Na_(3)PO_(4) D. After titration the final solution contains 0.0150 mol of H_(2)O.
Answer
4.7
(263 Votes)
Rowan
Master · Tutor for 5 years
Answer
B.
Explanation
This question pertains to the field of Acid-Base Titration in Chemistry. In a titration, one solution (the titrant) is added to another solution until the chemical reaction between the two solutes is complete. The point at which the reaction is complete is called the endpoint.Let's look at the equation: `H3PO4(aq) + 3 NaOH(aq) → Na3PO4(aq) + 3 H2O(l)`. Here, one mole of H3PO4 reacts with three moles of NaOH to completely neutralize it. This is the stoichiometric relationship between H3PO4 and NaOH. Furthermore, we're given the respective concentrations and volume of H3PO4(aq) involved, and the concentration of NaOH(aq).1. Calculate the number of moles of H3PO4(aq) involved. This is done by using the concentration and volume of H3PO4(aq). Moles = concentration × volume in dm^3. But since the provided volume is in cm^3, it's first converted to dm^3 as (cm^3 ÷ 1000).Number of moles of H3PO4(aq) = 0.200 mol/dm^3 * (25.0 cm3 ÷ 1000) = 0.005 mol.2. Go to the stoichiometric relationship to deduce the volume of NaOH(aq) required. Since it's at 3-fold compared to H3PO4(aq), the number of moles of NaOH needed is 3 times the amount of H3PO4(aq).Moles of NaOH needed = 3 × moles of H3PO4 = 3 × 0.005 mol = 0.015 mol.3. Calculate the end point by figuring out the volume of NaOH(aq) needed to neutralize H3PO4(aq). This involves the number of moles calculated (step 2) and the concentration of NaOH supplied. Volume = moles ÷ concentration.Volume of NaOH(aq) = 0.015 mol ÷ 0.600 mol/dm^3 = 0.025 dm^3.This volume is expressed in cm^3 to match the options and information given, hence,Volume of NaOH(aq) in cm^3 = 0.025 dm^3 * 1000 = 75 cm^3.Consequently, Option B is the correct statement, given that the endpoint or equivalence point of the titration occurs when 75.00 cm^3 of NaOH(aq) has been added.