Question

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

Answer 1

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