A student is given a solution of acetic acid (HC2H3O2) with an unknown concentra

Question

A student is given a solution of acetic acid (HC2H3O2) with an unknown concentration. To determine the concentration, the student titrates 25.00 ml of the acetic acid solution with 0.250 M NaOH. 29.7 ml of the NaOH is needed to complete the titration.

a.) what is the concentration (in molarity) of the acetic acid solution?

A student needs to make 750 ml of a 0.075 M nitric acid solution (HNO3, a strong acid) from a 2.5 M stock solution.

a.) what volume of the stock solution should the student use?

b.) what should the pH, pOH, and [OH^-] be in this solution if made correctly?

Explanation / Answer

At equivalence point, no. of moles of acetic acid = no. of moles of NaOH.

no. of moles of NaOH = molarity x volume in liters = 0.250M x 0.0297L = 7.425E-3moles

so, molarity of acetic acid = no. of moles / volume in liters = 7.425E-3/0.025L = 0.297M

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