Brand _____ Volume 5.0 mL (% on label) _____% Molarity (M) of NaOH _____ Initial

Question

Brand _____ Volume 5.0 mL (% on label) _____% Molarity (M) of NaOH _____ Initial NaOH level in buret Final NaOH level in buret Volume (ml of NaOH used Average volume (mL) Average volume in liters (L) Moles of NaOH used in titration (Show calculations.)_____ moles NaOH Moles of HC_2H_3O_2 neutralized by NaOH_____ mole HC_2H_3O_2 Molarity of HC_2H_3O_2 (Show calculations.) _____ M HC_2H_3O_2 Grams of HC_2H_3O_2 (Show calculations.) _____ g HC_2H_3O_2 Percent (m/v) HC_2H_3O_2 in vinegar (Show calculations.) _____ HC_2H_3O_2

Explanation / Answer

Q8

moles of NaOH = MV = 0.1*(44.06*10^-3) = 0.004406 moles of NaOH

Q9

1 mol of NaOH neutralizes 1 mol of acid

so

0.004406 mol of NaOH = 0.004406 mol of HC2H3O2

Q10

M = mol/V = 0.004406 /(5*10^-3) = 0.8812 M of acid

Q11

Mass in grams:

mass = mol*MW = 0.004406 *60 = 0.26436 g of acid are present

Q12

m/v% = mass of acid / total volume = 0.26436 g / 5 mL *100% = 5.2872 % m/v

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