...o AT&T; 1:17 PM session.masteringchemistry.com C Calculate the pll values and
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...o AT&T; 1:17 PM session.masteringchemistry.com C Calculate the pll values and draw the titration curve for the titration of 500 mL of O.010 M acetic acid (x 4.76) with 0.010 MKon Part A Calculate the pH of the solution after 0ml. of the titrant have been added Express your answer using two decimal places plt Submit My Answers Give Up Part B Caloulate the pli of the solution after 250 l of the titrant have been added Express your your answer using two decimal places plt Submit My Answers Give Up Part C Calculate the jll of the solution after 490l of the titrant have been added Express your answer using two decimal places pH Submit My Answers Give Up Part D Calculate the pH of the solution after 500 mL of the titrant have been added Express your answer using two decimal places. plt Submit My Answers Give Up Part E Calculate the pH of the solution after 510 L of the titrant have been added Express your answer using two decimal places.Explanation / Answer
2.8-A-pH= pKa + log [salt] / [acid]
mole of acetic acid = 100x10^-3 x1 = 0.1mol
mole of sodium accetate=100x10^-3x0.5= 0.05mol
pH=4.76+ log[0.1]/[0.05] = 4.76+ log 2 = 4.76+0.301= 5.061
B-pKa=2.14
mole of acid =250 x10^-3x0.3=0.075mol
mole of salt = 250x10^-3x0.8= 0.2 mol
pH =pKa + log [salt]/[acid]
pH =2.14+ log[0.2]/[0.075]=2.14+0.426=2.566
a- when oml of titrant is added
pH= pKa= 4.76
b-mole of acid= 500x10^-3x 0.01=0.005mol
mole of KOH= 250x10^-3x0.01=0.0025mol
pH = pKa + log base/acid= 4.76 -0.3=4.46
c-mole of KOH= 490x10^-3x0.01= 0.0049mol
pH = pKa + log base/acid
pH= 4.76 + log [0.0049] /[0.005]=4.76-0.0087=4.751
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