strong acid strong base titration curve, Schemes and Mind Maps of Chemistry

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The equivalence point of the titration is the point at which exactly enough titrant has been added to react with all of the substance being titrated with no titrant left over. In other words, at the equivalence point, the number of moles of titrant added so far corresponds exactly to the number of moles of substance being titrated according to the reaction stoichiometry. (In an acidbase titration, there is a 1:1 acid:base stoichiometry, so the equivalence point is the point where the moles of titrant added equals the moles of substance initially in the solution being titrated.) Here is an example of a titration curve, produced when a strong base is added to a strong acid. This curve shows how pH varies as 0.100 M NaOH is added to 50. mL of 0.100 M HCl.

Calculate the pH at any point, including the

equivalence point, in an acidbase titration.

At the equivalence point, the pH = 7.00 for strong acidstrong base titrations. However, in other types of titrations, this is not the case The original number of moles of H

in the solution is:

What is the pH when 49.00 mL of 0.100 M NaOH solution have been added to 50.00 mL of 0.100 M HCl solution? Because it is a strong acidbase reaction, the reaction simplifies to: H

(aq) + OH (aq) H 2 O (l) The original number of moles of H

in the solution is:

What is the pH when 49.00 mL of 0.100 M NaOH solution have been added to 50.00 mL of 0.100 M HCl solution? Because it is a strong acidbase reaction, the reaction simplifies to: H

(aq) + OH (aq) H 2 O (l) The original number of moles of H

in the solution is: 50.00 x 10 3 L x 0.100 M HCl = 5.00 x 10 3 moles The number of moles of OH added is : 49.00 x 10 3 L x 0.100 M OH = 4.90 x 10 3 moles

What is the pH when 49.00 mL of 0.100 M NaOH solution have been added to 50.00 mL of 0.100 M HCl solution? Because it is a strong acidbase reaction, the reaction simplifies to: H

(aq) + OH (aq) H 2 O (l) The original number of moles of H

in the solution is: 50.00 x 10 3 L x 0.100 M HCl = 5.00 x 10 3 moles The number of moles of OH added is : 49.00 x 10 3 L x 0.100 M OH = 4.90 x 10 3 moles Thus there remains: (5.00 x 10 3 ) (4.90 x 10 3 ) = 1.0 x 10 4 moles H

(aq) The total volume of solution is 0.04900 L + 0.05000 L = 0.09900 L [H

] = {1.0 x 10 4 moles / 0.09900 L } = 1.0 x 10 3 M pH = 3.