Initial NaOH reading (Interpolate to 0.1 mL) Final NaOH reading (Interpolate to
ID: 938118 • Letter: I
Question
Initial NaOH reading
(Interpolate to 0.1 mL)
Final NaOH reading
(Interpolate to 0.1 mL)
Volume of NaOH used
Trial 1
9.0 mL
.63 mL
8.37 mL
Trial 2
9.8 mL
1.58 mL
8.22 mL
Trial 3
9.98 mL
1.61 mL
8.37 mL
Average volume of NaOH used: 8.32 mL
How do I calculate the normality of the vinegar using this equation? Na = Nb (volume b)/(volumea ) when using 5 mL of vinegar?
How to calculate the mass of the acetic acid in grams using the previously given equation. Massa = (Na) (GMWa)
Initial NaOH reading
(Interpolate to 0.1 mL)
Final NaOH reading
(Interpolate to 0.1 mL)
Volume of NaOH used
Trial 1
9.0 mL
.63 mL
8.37 mL
Trial 2
9.8 mL
1.58 mL
8.22 mL
Trial 3
9.98 mL
1.61 mL
8.37 mL
Average volume of NaOH used: 8.32 mL
Explanation / Answer
Normality of Vinegar can be calculated as, N1V1=N2V2
(0.5 M NaOH) x (8.32 mL NaOH) / (5 mL vinegar) = 0.832 M = 0.645 N vinegar
(0.832 mol/L) x ( 60.05196 g acetic acid/mol) = 49.96 g/L acetic acid