HC,H7o, (ag)+ NaOH(aq) NaC,H70, (ag)+ H2O(aq) First, we cailculate the number of
ID: 809059 • Letter: H
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HC,H7o, (ag)+ NaOH(aq) NaC,H70, (ag)+ H2O(aq) First, we cailculate the number of moles of 200 g of butyric acid and o.50 9 NaOH 2.00 g HC,H,O,x I mol HC.H.o, 0.50 g NaOH x 39,998 g NaOH -Innol HCM,02--= 0.0227 mol HC,HR2 114 200P l mol Hearth 88.104 g HC H,0 #0.0227 mol HC,HR2 2.00 g HC,H,0.* 0.50 g NaOHx mol NaOH 0.50 g NaoHx I mol NaoH I mol NaOH-0.0125 mol NaOH 39.998 g NaOH Now, construct the ICE table. HC,H,02 (aq) + NaOH (aq)-= NaC,H,02 (aq) + H2O(ag) Initial(mol).0022700125 0.0227 00125 Initial(mol): Change(mol):0.012500125 Equilibrium(mol): 0.0102 Provide feedback (0) 0.00 +0.0125 0.0125 Step 2 of 2 Calculate the pH of the solution using Henderson-Hasselbalch equation.Explanation / Answer
Here NaOH is limiting, thus the number of mols of formed as the product is equals to the number of mols of NaOH.
Thus , at equilibrium both acid and its conjugate base(salt ) both can be existed.
So, according to Henderson's equation -
pH = pKa + log [NaC4H7O2] / [HC4H7O2]
pH = pKa + log (0.0125 / 0.0102 )
= pKa + 0.088
Thus, simply substitute pKa value in the above relation to get the pH.
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