Question

1. a student was performing a reaction of neutralization of baking soda with vinegar (the same reaction you will be performing during this lab). the student placed 2.95 g of baking soda in a large cup and kept adding vinegar until there were no more bubbles. the diagram below shows the initial measured numbers, and also what happened during the experiment. calculate the mass % of acetic acid in vinegar (filling in the blanks below will guide you). look through the diagram left to right, step by step. notice how half way through the procedure there is no more undissolved baking soda left (unreacted nahco₃ is in solution at that stage). if you are not quite sure what is happening in it, re - read the “lab procedure”. a) mass nahco₃ = ________ g (0.5 pt) b) density vinegar = 26.25 g / 25.0 ml = ______ g/ml (0.5 pt) c) total volume vinegar = __+__+__ = __ ml (step 9 in the lab procedure) (1 pt) d) moles nahco₃ = g / 84.007 g/mol = ______ moles (1 pt) e) moles ch₃cooh = ______ moles (1 pt) f) mass ch₃cooh = moles ch₃cooh × molar mass = ______ moles×60.052 g/mol = __g (1 pt) g) mass vinegar = total vol. vinegar×density of vinegar = __ml × __g/ml = __g (1 pt) h) mass % of acetic acid in vinegar = (mass acetic acid / mass vinegar)×100% = __g / __g×100% = ____% sig figs

Explanation:

Step1: Identify given mass of $NaHCO_3$

The mass of $NaHCO_3$ is given as $2.95$ g. So for part a), Mass $NaHCO_3 = 2.95$ g.

Step2: Calculate density of vinegar

Density of vinegar $
ho=\frac{m}{V}$, where $m = 26.25$ g and $V = 25.0$ mL. So $
ho=\frac{26.25}{25.0}=1.05$ g/mL for part b).

Step3: Calculate total volume of vinegar

The volumes of vinegar added are $25.0$ mL, $15$ mL, $12$ mL and $1$ mL. Total volume $V_{total}=25.0 + 15+12 + 1=53.0$ mL for part c).

Step4: Calculate moles of $NaHCO_3$

Moles of $NaHCO_3$, $n=\frac{m}{M}$, where $m = 2.95$ g and $M = 84.007$ g/mol. So $n=\frac{2.95}{84.007}\approx0.0351$ moles for part d).

Step5: Determine moles of $CH_3COOH$

The balanced chemical - equation for the reaction between $NaHCO_3$ and $CH_3COOH$ is $NaHCO_3+CH_3COOH = CH_3COONa + H_2O+CO_2\uparrow$. The mole ratio of $NaHCO_3$ to $CH_3COOH$ is $1:1$. So moles of $CH_3COOH = 0.0351$ moles for part e).

Step6: Calculate mass of $CH_3COOH$

Mass of $CH_3COOH$, $m=n\times M$, where $n = 0.0351$ moles and $M = 60.052$ g/mol. So $m = 0.0351\times60.052\approx2.11$ g for part f).

Step7: Calculate mass of vinegar

Mass of vinegar, $m = V\times
ho$, where $V = 53.0$ mL and $
ho = 1.05$ g/mL. So $m = 53.0\times1.05 = 55.65$ g for part g).

Step8: Calculate mass - percentage of acetic acid in vinegar

Mass - percentage of acetic acid in vinegar, $\%=\frac{\text{mass of acetic acid}}{\text{mass of vinegar}}\times100\%=\frac{2.11}{55.65}\times100\%\approx3.79\%$ for part h).

Answer:

a) $2.95$
b) $1.05$
c) $25.0+15 + 12+1 = 53.0$
d) $\frac{2.95}{84.007}\approx0.0351$
e) $0.0351$
f) $0.0351\times60.052\approx2.11$
g) $53.0\times1.05 = 55.65$
h) $\frac{2.11}{55.65}\times100\%\approx3.79$