29 Na te Lab Section Prelaboratory Problems- Experiment 3 - Acids and Bases: Ana

Question

29 Na te Lab Section Prelaboratory Problems- Experiment 3 - Acids and Bases: Analysis 1.° The titration of 25.00 mL of a sulfuric acid solution of unknown concentration requires 31.22 mL of a 0.1234 M NaOH solution. What is the concentration of the sulfuric acid solution? 2. 10.00 mL of vinegar (mass 10.05 g) requires 16.28 mL of 0.5120 M NaOH to reach the end point. Calculate the molarity and mass percent of the acetic acid in the vinegar. 3. A 0.1936 g sample of an unknown monoprotic acid requires 15.56 mL of 0.1020 M NaOH solution to reach the end point. What is the molecular mass of the acid? 2015 Cengage Leaming All Rights Reserved May not be xcanned, copied or duplicated, or posted to a publidy accessitle websle,in whole or in pe

Explanation / Answer

1) Volume of NaOH V = 31.22 ml = 0.03122 L

Volume of  sulphuric acid = 25.00ml = 0.025 L

H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

means 1 mole of sulphuric acid requires 2 moles of NaOH

moles of NaOH = M*V

= 0.1234*0.03122

= 3.853 *10^-3

moles of  H2SO4 = moles of NaOH / 2

=  3.853 *10^-3/2

= 1.926 *10^-3 Mol

concentration of H2SO4 ( sulphuric acid) = Moles/ V   

= 1.926*10^-3 / 0.02500

= 0.07705M

2) CH3COOH ------------> CH3COO-+ H+

gives one H+ ion so this is mono acidic NaOH is monobasic

so N1=N2

moles of base NaOH = 0.5120 mol/ltr * 16.28 !ltr/1000ml

=0.00833 moles

M1V1 = M2V2

V1 = 10 ml

M2 = 0.5120

V2 = 16.28

M1 = M2V2/V1

= 0.833 M

mass percent

molecular mass of acetic acid 60.05 gm/mol

mass of acetic acid = molecular mass *no.of moles

= 60.05 * 0.00833

= 0.5002 g

mass percent = 0.5002/ 10.05 = 4.35%

3) mass of monoprotic acid = 0.1936 g

volume of NaOH = 15.56 ml

molarity of NaOH = 0.1020M

mol OH- = .1020 * .01556 = 0.00158 mol NaOH

at end point one mole of acid reacts with 1 o of base so 0.00158 moles of NaOH reacts with 0.00158 moles of H+

molecular mass of acid = 0.1936/0.00158 = 122.5 g

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