Chapter 10: Q. 10.96 (page 360)

A 10.0-mL sample of vinegar, which is an aqueous solution of acetic acid, ${HC}_{2} H_{3} O_{2}$ , requires $16.5 mL$ of a $0.500 MNaOH$ solution to reach the endpoint in a titration. What is the molarity of the acetic acid solution? $(10.7)$
${HC}_{2} H_{3} O_{2} (a q) + NaOH (a q) ⟶ H_{2} O (f) + {NaC}_{2} H_{3} O_{2} (a q)$

Short Answer

Expert verified

Molarity ${CH}_{3} COOH = \frac{0.0083 {moleCH}_{3} COOH}{0.010 L}$

$= 0.830 {MCHCOOH}_{3} OOH$

Step by step solution

Given information

That are given numbers are calculated by:

Equalities /conversion factors

Explanation

Step 1 Given $16.5 ml$ of $0.500 MNaOH; 10.0 {mlCH}_{3} COOH = 0.010 {LCH}_{3} COOH$

Need molarity of ${CH}_{3} COOH$ .

Step 2 Plan

Metric Molarity mole Divide by

localid="1654585928067" $16.5 mlL$ moles $NaOH$ mole ${CH}_{3} COOHM$

Factor factor factor liters

Step 3 Equalities/conversion factors

$\begin{aligned} 1 L NaOH = 0.500 moleNaOH \\ \frac{1 L}{0.500 mole} = \frac{0.500 mole}{1 L} \\ 1 L NaOH = 1000 mlNaOH \\ \frac{1 L}{1000 ml} = \frac{1000 ml}{1 L} \end{aligned}$

Given information

That are given equations are:

Explanation

Step 2 of 2

l mole $NaOH = 1 {moleCHCOOH}_{3} COOH$

Step 4 Set up problem

$16.5 mLNaOH \times \frac{1 L SaOH}{1000 mLNaOH} \times \frac{0.500 mole . AaOH}{1 LNeOH} \times \frac{1 {moleCHCOOH}_{3}}{1 moleNaOH}$

$= 0.0083 {moleCHCHOH}_{3} OOH$

Molarity ${CH}_{3} COOH = \frac{0.0083 {moleCH}_{3} COOH}{0.010 L}$

$= 0.830 {MCH}_{3} COOH$

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A 10.0-mL sample of vinegar, which is an aqueous solution of acetic acid,HC2H3O2, requires 16.5mLof a 0.500MNaOH solution to reach the endpoint in a titration. What is the molarity of the acetic acid solution? (10.7)HC2H3O2(aq)+NaOH(aq)⟶H2O(f)+NaC2H3O2(aq)

Short Answer

Step by step solution

Given information

Explanation

Given information

Explanation

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Most popular questions from this chapter

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