Determination of the Molar Mass of the unknown acid Average molarity of NaOh sol

Question

Determination of the Molar Mass of the unknown acid

Average molarity of NaOh solution is 0.0997 M

Volume of NaOH used was 40.00 mL for trial 1 and 40.58 mL for trial 2.

Initial reading on NaOH buret trial 1 and trial 2 was 50.00 mL

Final reading on NaOH buret for trial 1 was 10.00 mL and 9.42 mL for trial 2.

A. Number of moles NaOH= (V(NaOH) * M(NaOH)) / 1000. calculate for trial 1 and 2_______

B. Number of moles H+ in sample. This will equal the number of moles of OH- in the NaOH. calculate for trial 1 and 2_________

C. MM=number of grams acid / number moles H+. calculate for trial 1 and 2_______________g

Explanation / Answer

we know that

moles = molarity x volume (ml) /1000

trail 1 :

moles of NaOh = 0.0997 x 40 /1000 = 3.988 x 10-3

trail 2 :

moles of NaOH used = 0.0997 x 40.58 /1000 = 4.0458 x 10-3

B)

NaOH ----> Na+ + OH-

from the above reaction

moles of OH- = moles of NaOH

also

H+ + OH- ---> H20

so

moles of H+ = moles of OH-

so

moles of H+ = moles of NaOH

trail 1 :

moles of H+ = 3.988 x 10-3

trail 2 :

moles of H+ = 4.0458 x 10-3

C)

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

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