1.(5 points) A 0.6251 gram sample of acid potassium phthalate (GFW = 204.23) req

Question

1.(5 points) A 0.6251 gram sample of acid potassium phthalate (GFW = 204.23) requires 18.05 mL of a NaOH solution to completely neutralize it using phenolphthalein as an indicator. What is the molarity of the NaOH solution? Show work

2.(5 points) A 10.00 mL sample a of an acetic acid , HC2H3O2 solution requires 25.35 mL of a 0.2255 M NaOH solution to completely neutralize it using a phenolphthalein indicator. Calculate the molarity of the acetic acid solution.
HC2H3O2 + NaOH ! NaC2H3O2 + H2O Show work

Explanation / Answer

Answer:

1) Given weight of acid potassium phthalate=0.6251 g and molecular weight=204.23 g/mol,

Then number of moles of acid= weight/molecular weight

=0.6251g/204.23 g/mol=3.06x10^-3 moles.

Mole ratio between acid to NaOH is 1:1,

andvolume of NaOH=18.05 mL=0.01805 L.

So concentration of NaOH=moles/L=3.06x10^-3 moles/0.01805 L

[NaOH]=0.1695 M.

2) Given acetic acid volume=10 mL=0.010 L,

and NaOH volume=25.35 mL, Molarity=0.2255 M,

So moles of NaOH=Molarity x volume=0.2255 moles/Lx0.02535 L=5.716x10^-3 moles.

The balanced equation HC2H3O2+NaOH----->NaC2H3O2+H2O

So moles of acid =moles of NaOH

Molarity of acid=moles/volume (L)=5.716x10^-3 moles/0.010L

Molarity of acid=0.5716 M.

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.