Question

please help with my pre lab

Pre-Lab Questions: 1. What is the definition of an 'equivalence point' in an acid/base titration? (1 point) 2. In part one of the experiment, you will prepare the acid solutions being titrated from a stock solution. Describe how you will accurately prepare 10.00 mL of 0.100 M HCl solution using a 1.00 M HCI stock solution. In your response to this question, be very specific about the quantities of stock solution and deionized water to be used in the dilution and the specific types of glassware you will use. (2 points) 3. If 15.0 mL of 0.125 M phosphoric acid is titrated with 5.100 M NaOH, what volume of the titrant (in mL) must be added to completely neutralize the acid? Show all of your work (including the chemical equation). (1 point)

additional information

Answer 1

Answer (1) In acid base titrations, during experiment we keep on adding one to the other i.e. an acid to base or vice-versa. At the point where enough acid/reactant has been added which can neutralise the base/analyte present in the container is called the equivalence point. In other words at this point the moles of the reactant added are equal to moles of the analyte present.

Answer (2) Molarity of stock HCl solution= 1M

Volume of this solution = Unknown ( to be calculated)

Molarity of solution to be prepared = 0.1M

Volume of this solution to be prepared = 10mL

using the formula:

M₁*V₁=M₂*V₂

Putting in the values now we get

1*V= 0.1*10

V=1mL

Thus to prepare 0.1M HCl solution from 1M stock solution, 1mL of the 1M stock solution should be added to 9mL distilled water.

Glasswares used: Burette, pipette, stand, beakers, funnel, glass rod

Answer (c) Molarity of phosphoric acid= 0.125M

Volume of phosphoric acid = 15mL

Molarity of NaOH solution = 0.100 M

Volume of NaOH solution to neutralise acid = unknown ( to be calculated)

Using the formula:

M₁*V₁=M₂*V₂

_{0.125 * 15 = 0.100 * V}

V = 18.75 mL

Thus the amount of base required to neutralise the acid is 18.75 mL.

Simple acid base neutralisation reaction:

H⁺ + OH^- ---> H₂O