Question

Determine the % of acetic acid (MM = 60.05 g/mol) in a 2.00 mL vinegar sample that has a mass of 1.98 g to two decimal places if 15.53 mL of a 0.088 M NaOH solution were used to titrate the vinegar sample.

Answer 1

1) Balance the chemical equation

`CH_3COOH+NaOH\rightarrow NaCH_3COO+H_2O`

2) Moles of NaOH in the reaction

Convert mL into L

`L=15.53mL\cdot(1L)/(1000mL)=0.01553L`

Find moles of NaOH

`M=\frac{\text{moles of solute}}{\text{liters of solution}}`

Plug in values and solve for moles

`0.088M=\frac{\text{moles of NaOH}}{0.01553L}`
`\text{moles of NaOH=0.088M}\cdot0.01553L=0.00136664\text{ mol NaOH}`

3) Moles of acetic acid that reacted with 0.00136664 mol NaOH

Molar ratio

1 mol NaOH: 1 mol CH3COOH

`\text{mol CH}_3COOH=0.00136664\text{ mol NaOH}\cdot\frac{1molCH_3COOH}{1\text{ mol NaOH}}=0.00136664molCH_3COOH`

4) Convert moles of CH3COOH into grams of CH3COOH

The molar mass of CH3COOH is 60.05 g/mol

`gCH_3COOH=0.00136664molCH_3COOH\cdot(60.05gCH_3COOH)/(1molCH_3COOH)=0.0820667gCH_3COOH`

The mass of acetic acid is 0.0820667g

5) Percent of acetic acid

`\text{Mass}\%=\frac{\text{grams of solute}}{\text{grams of solution}}\cdot100`
`\text{Mass}\%=\frac{0.0820667gCH_3COOH}{1.98\text{ g of solution}}\cdot100=4.14\%`

The percent of acetic acid by mass is 4.14%.

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