# Return to Answer

 7 replaced http://chemistry.stackexchange.com/ with https://chemistry.stackexchange.com/ edited Apr 13 '17 at 12:57 If the initial solution is just acetic acid and water, and the pH is 3.97, this means that it is a very dilute solution, less than 0.001M acetic acid. See http://depts.washington.edu/chem/facilserv/lecturedemo/pHofAceticAcid-UWDept.ofChemistry.html Therefore, of the 0.025 moles of OH-, almost all (more than .024 moles) will remain unreacted. pH = 14 - pOH pH = 14 + log(0.025) = 12.4 In a different situation, where you added less strong base than weak acid, the Henderson-Hasselbach equation would be helpful, but it is not helpful in this example. If the initial solution is just acetic acid and water, and the pH is 3.97, this means that it is a very dilute solution, less than 0.001M acetic acid. See http://depts.washington.edu/chem/facilserv/lecturedemo/pHofAceticAcid-UWDept.ofChemistry.html Therefore, of the 0.025 moles of OH-, almost all (more than .024 moles) will remain unreacted. pH = 14 - pOH pH = 14 + log(0.025) = 12.4 In a different situation, where you added less strong base than weak acid, the Henderson-Hasselbach equation would be helpful, but it is not helpful in this example. If the initial solution is just acetic acid and water, and the pH is 3.97, this means that it is a very dilute solution, less than 0.001M acetic acid. See http://depts.washington.edu/chem/facilserv/lecturedemo/pHofAceticAcid-UWDept.ofChemistry.html Therefore, of the 0.025 moles of OH-, almost all (more than .024 moles) will remain unreacted. pH = 14 - pOH pH = 14 + log(0.025) = 12.4 In a different situation, where you added less strong base than weak acid, the Henderson-Hasselbach equation would be helpful, but it is not helpful in this example. 6 added 167 characters in body edited Feb 28 '15 at 17:59 DavePhD 36k22 gold badges6363 silver badges149149 bronze badges If the initial solution is just acetic acid and water, and the pH is 3.97, this means that it is a very dilute solution, less than 0.001M acetic acid. See http://depts.washington.edu/chem/facilserv/lecturedemo/pHofAceticAcid-UWDept.ofChemistry.html Therefore, of the 0.025 moles of OH-, almost all (more than .024 moles) will remain unreacted. pH = 14 - pOH pH = 14 + log(0.025) = 12.4 In a different situation, where you added less strong base than weak acid, the Henderson-Hasselbach equation would be helpful, but it is not helpful in this example. If the initial solution is just acetic acid and water, and the pH is 3.97, this means that it is a very dilute solution, less than 0.001M acetic acid. See http://depts.washington.edu/chem/facilserv/lecturedemo/pHofAceticAcid-UWDept.ofChemistry.html Therefore, of the 0.025 moles of OH-, almost all (more than .024 moles) will remain unreacted. pH = 14 - pOH pH = 14 + log(0.025) = 12.4 If the initial solution is just acetic acid and water, and the pH is 3.97, this means that it is a very dilute solution, less than 0.001M acetic acid. See http://depts.washington.edu/chem/facilserv/lecturedemo/pHofAceticAcid-UWDept.ofChemistry.html Therefore, of the 0.025 moles of OH-, almost all (more than .024 moles) will remain unreacted. pH = 14 - pOH pH = 14 + log(0.025) = 12.4 In a different situation, where you added less strong base than weak acid, the Henderson-Hasselbach equation would be helpful, but it is not helpful in this example. 5 added 128 characters in body edited Feb 28 '15 at 17:45 DavePhD 36k22 gold badges6363 silver badges149149 bronze badges If the initial solution is just acetic acid and water, and the pH is 3.97, this means that it is a very dilute solution, less than 0.001M acetic acid. See http://depts.washington.edu/chem/facilserv/lecturedemo/pHofAceticAcid-UWDept.ofChemistry.html Therefore, of the 0.025 moles of OH-, almost all (more than .024 moles) will remain unreacted. pH = 14 - pOH pH = 14 + log(0.025) = 12.4 If the initial solution is just acetic acid and water, and the pH is 3.97, this means that it is a very dilute solution, less than 0.001M acetic acid. See http://depts.washington.edu/chem/facilserv/lecturedemo/pHofAceticAcid-UWDept.ofChemistry.html Therefore, of the 0.025 moles of OH-, almost all (more than .024 moles) will remain unreacted. pH = 14 - pOH pH = 14 + log(0.025) = 12.4 If the initial solution is just acetic acid and water, and the pH is 3.97, this means that it is a very dilute solution, less than 0.001M acetic acid. See http://depts.washington.edu/chem/facilserv/lecturedemo/pHofAceticAcid-UWDept.ofChemistry.html Therefore, of the 0.025 moles of OH-, almost all (more than .024 moles) will remain unreacted. pH = 14 - pOH pH = 14 + log(0.025) = 12.4 Post Undeleted by DavePhD occurred Feb 28 '15 at 17:38 4 added 109 characters in body edited Feb 28 '15 at 17:38 DavePhD 36k22 gold badges6363 silver badges149149 bronze badges 3 added 109 characters in body edited Feb 28 '15 at 17:33 DavePhD 36k22 gold badges6363 silver badges149149 bronze badges Post Deleted by DavePhD occurred Feb 28 '15 at 17:29 2 added 109 characters in body edited Feb 28 '15 at 17:26 DavePhD 36k22 gold badges6363 silver badges149149 bronze badges 1 answered Feb 28 '15 at 17:20 DavePhD 36k22 gold badges6363 silver badges149149 bronze badges