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Problem 14.17 Acetic acid, CH3CO2H, is the solute that gives vinegar its characteristic odor and sour...

Problem 14.17 Acetic acid, CH3CO2H, is the solute that gives vinegar its characteristic odor and sour taste. Part A Calculate the pH in 1.80 M CH3CO2H. Express your answer using two decimal places. pH = 2.31 SubmitMy AnswersGive Up Incorrect; Try Again; 2 attempts remaining Part B Calculate the concentrations of all species present (H3O+, CH3CO2−, CH3CO2H, and OH−) in 1.80 M CH3CO2H. Express your answers using two significant figures separated by commas. Enter your answers in order listed in the question. [H3O+], [CH3CO2−], [CH3CO2H], [OH−] = M SubmitMy AnswersGive Up Part C Calculate the pH in 0.183 M CH3CO2H. Express your answer using two decimal places. pH = SubmitMy AnswersGive Up Part D Calculate the concentrations of all species present (H3O+, CH3CO2−, CH3CO2H, and OH−) in 0.183 M CH3CO2H. Express your answers using two significant figures separated by commas. Enter your answers in order listed in the question. H3O+, CH3CO2−, CH3CO2H, OH− = M SubmitMy AnswersGive Up

Solutions

Expert Solution

acetic acid ionizes as CH3COOH+ H2O ----->CH3COO- + H3O+

Ka= [CH3COO-] [H3O+]/[CH3COOH] = 1.8*10-5,

ICE table is prepares

Compound                                           initial (M)                            change                           equilibrium

CH3COOH                                           1.8                                         -x                                    1.8-x

CH3COO-                                              0                                           x                                         x

H3O+                                                     0                                          x                                         x

Ka= x2/(1.8-x )= 1.8*10-5, when solved using excel, x= 0.0057, [H3O+]=0.0057M, pH= -log [H3O+]= 2.24

pOH= 14-2.24= 11.76, [OH-]= 10(-11.76)= 1.74*10-12, [CH3COO-]=0.0057M, [CH3COOH] = 1.8-0.0057= 1.7943M

2. when the concentration of acetic acid is 0.183M,

Ka =x2/(0.183-x)= 1.8*10-5, when sovled using excel, x= 18.1*10-4, [H3O+]= [CH3COO-]= 18.1*10-4 M,

pH= 2.74, pOH= 14-2.74= 11.26, [OH-]= 5.525*10-12, [CH3COOH]=0.183-18.1*10-4= 0.18119M


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