Question

acetic acid ($ce{hc_{2}h_{3}o_{2}}$) is an important ingredient of vinegar. a sample of 50.0 ml of a commercial vinegar is titrated against a 1.00 m naoh solution. what is the concentration (in m) of acetic acid present in the vinegar if 5.35 ml of the base is needed for the titration? be sure your answer has the correct number of significant digits.

Explanation:

Step1: Recall the titration formula

For a neutralization reaction between an acid (acetic acid, \(HC_2H_3O_2\)) and a base (NaOH), the moles of acid equal the moles of base at the equivalence point. The formula is \(M_1V_1 = M_2V_2\), where \(M_1\) and \(V_1\) are the molarity and volume of the acid, and \(M_2\) and \(V_2\) are the molarity and volume of the base.

Step2: Identify the given values

We know that \(M_2 = 1.00\ M\), \(V_2 = 5.35\ mL\), and \(V_1 = 50.0\ mL\). We need to find \(M_1\).

Step3: Rearrange the formula to solve for \(M_1\)

From \(M_1V_1 = M_2V_2\), we can rearrange it to \(M_1=\frac{M_2V_2}{V_1}\).

Step4: Substitute the values into the formula

Substitute \(M_2 = 1.00\ M\), \(V_2 = 5.35\ mL\), and \(V_1 = 50.0\ mL\) into the formula: \(M_1=\frac{1.00\ M\times5.35\ mL}{50.0\ mL}\).

Step5: Calculate the result

First, calculate the numerator: \(1.00\ M\times5.35\ mL = 5.35\ mmol\) (since \(M=\frac{mmol}{mL}\) for volume in mL). Then divide by the volume of the acid: \(\frac{5.35\ mmol}{50.0\ mL}=0.107\ M\).

Answer:

\(0.107\)