Titrations are acid-base reactions in which the amount of acid and base in solution are known quite precisely. The specific case wherethe acid and base have been added in equivalent molar amounts, is called the equivalence point. At this point,mol acid = mol base%3DAt all other points in a titration, either the base is the limiting reactant (and there is excess acid) or the acid is the limiting reactant (andthere is excess base). So, except at the equivalence point, titration problems are limiting reactant problems, which means we can dothem- a lot of them!One additional point: when there is left over strong acid (acid in excess), 100% of that excess acid is ionized to become H* ions:[strong acid] = [H*] and pH = -log[H"]Something similar happens for strong bases.1. Calculate the pH of the solution after the following amounts of 0.250 M NaOH have been added to 50.0 mL of 0.200 M HCI. Showall your work for each calculation (including the volumes for which the pH is given). Feel free to attach additional pages to yoursubmission if needed.mL of NaOHpH0.00ANS: 0.69936.0039.7540.0040.2544.0066.00ANS: 12.752. The point at 40.00 mL of NaOH is a turning point in the calculations. How so? What should the pH be at/40.00 mL and why? Howabout after 40.00 mL?

HCl and NaOH react by neutralisation reaction HCl +NaOH→NaCl +H2O. Moles of HCl present =Molarity…

Answered: 1. Calculate the pH of the solution…

Chemistry

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

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Question

Titrations are acid-base reactions in which the amount of acid and base in solution are known quite precisely. The specific case where
the acid and base have been added in equivalent molar amounts, is called the equivalence point. At this point,
mol acid = mol base
%3D
At all other points in a titration, either the base is the limiting reactant (and there is excess acid) or the acid is the limiting reactant (and
there is excess base). So, except at the equivalence point, titration problems are limiting reactant problems, which means we can do
them- a lot of them!
One additional point: when there is left over strong acid (acid in excess), 100% of that excess acid is ionized to become H* ions:
[strong acid] = [H*] and pH = -log[H"]
Something similar happens for strong bases.
1. Calculate the pH of the solution after the following amounts of 0.250 M NaOH have been added to 50.0 mL of 0.200 M HCI. Show
all your work for each calculation (including the volumes for which the pH is given). Feel free to attach additional pages to your
submission if needed.
mL of NaOH
pH
0.00
ANS: 0.699
36.00
39.75
40.00
40.25
44.00
66.00
ANS: 12.75
2. The point at 40.00 mL of NaOH is a turning point in the calculations. How so? What should the pH be at/40.00 mL and why? How
about after 40.00 mL?

Expert Solution

Step 1

HCl and NaOH react by neutralisation reaction $H C l + N a O H \to N a C l + H 2 O$ .

Moles of HCl present

$= M o l a r i t y \times V o l u m e = 0.200 M \times 50 \times 10^{- 3} L = 0.01 m o l$

So at every point we will calculate the moles of 0.250M NaOH present and then decide whether H+ or OH- is in excess by subtracting the moles of HCl or NaOH whichever is higher from the another one. This will give excessive moles of HCl or NaOH present. Total volume present will be the sum of volume of HCl (50 ml) and NaOH at that point. Concentration of excessive H+ or OH- is given by

[H+/OH-] = (moles of excessive H+/OH-) /(sum of volume of HCl and NaOH)

When H+ is in excess we will calculate pH directly and when OH- will be in excess we will calculate pOH and then subtract it from 14 as pH +pOH =14.