A 0.1375 M solution of KOH is used to titrate 35.00 mL of 0.257 M HBr. ( Assume that volumes are additive) a. write a balanced net ionic equation for the reaction that takes place during titration b. what are the species present at the equivalence point? c. what volume of KOH is required to reach the equivalence point? d. what is the pH of the solution before any KOH is added?
A 0.1375 M solution of KOH is used to titrate 35.00 mL of 0.257 M HBr. ( Assume that volumes are additive) a. write a balanced net ionic equation for the reaction that takes place during titration b. what are the species present at the equivalence point? c. what volume of KOH is required to reach the equivalence point? d. what is the pH of the solution before any KOH is added?
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A 0.1375 M solution of KOH is used to titrate 35.00 mL of 0.257 M HBr. ( Assume that volumes are
additive)
a. write a balanced net ionic equation for the reaction that takes place during titration
b. what are the species present at the equivalence point?
c. what volume of KOH is required to reach the equivalence point?
d. what is the pH of the solution before any KOH is added?
Please refer to the pictures. Thank you so much.
![14
12
10
Equivalence
point
BB
(Neutral
FIGURE 14.5
A strong acid-strong base titration. The
curve represents the titration of 50.00 mL
of 1.000 M HCI with 1.000M NAOH.
The solution at the equivalence point is
neutral (pH = 7). The pH rises so rapidly
near the equivalence point that any of the
three indicators, methyl red (MR, end
point pH = 5), bromthymol blue (BB,
end polnt pH = 7), or phenolphthalein
(PP, end point pH = 9) can be used. OHT
pH)
MI
25
50
75
100
ml. NaOH added
Example: when 50.00 mL of 1.000 M HCI is titrated with 1.000 M NaOH, the pH increases. Find the pH of
b. 50.01
the solution after the following volumes of 1.000 M NaOH have been added.
a. 49.99 mL
ml
Solution: The number of moles of H* in 50.00 mL of 1.000 M HCI is
nH. = 50.00 x 10³ L x 1.000 mol H*/1L = 50.00 x 103 mol of H*
in each part of the problem, start 1. By calculating the number of moles of OH added then 2. Find the
number of moles of H* or OH in excess. Finally 3. Calculate [H*] and the pH.
a. (1) noh- = 49.99 x 10³ L x 1.000 mol OH/1L = 49.99 x 10³ mol of OH
(2) because H* and OH react in a 1:1 mole ratio,
NH+ reacted = 49.99 x 103 mol
nH+ excess = 50.00 x 103 mol – 49.99 x 103 mol = 0.01 x 103 = 1 x 105 mol
(3) the total volume is almost exactly 100 mL ( 50.00 mL + 49.99 mL):
[H] = 1 x 105 mol H*/1 x 101 = 1 x 104 M
pH = 4
b. (1) noH- = 50.01 x 10³L x 1.000 mol OH/ 1L = 50.01 x 10³ mol of OH-
(2) nH+ reacted = 50.00 x 10³ mol
noH- excess = 50.01 x 10³ mol - 50.00 x 10³ mol = 0.01 x 10³ mol = 1 x 105 mol
(3) [OH-] = 1 x 10$ mol OH-/1 x 101 L = 1x 104 M
pOH = 4.0; pH = 10.0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a57110d-a0e6-42a9-b8a3-769a5c198240%2F57114ced-7b21-4428-b935-3b3408996250%2Ff3uphg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:14
12
10
Equivalence
point
BB
(Neutral
FIGURE 14.5
A strong acid-strong base titration. The
curve represents the titration of 50.00 mL
of 1.000 M HCI with 1.000M NAOH.
The solution at the equivalence point is
neutral (pH = 7). The pH rises so rapidly
near the equivalence point that any of the
three indicators, methyl red (MR, end
point pH = 5), bromthymol blue (BB,
end polnt pH = 7), or phenolphthalein
(PP, end point pH = 9) can be used. OHT
pH)
MI
25
50
75
100
ml. NaOH added
Example: when 50.00 mL of 1.000 M HCI is titrated with 1.000 M NaOH, the pH increases. Find the pH of
b. 50.01
the solution after the following volumes of 1.000 M NaOH have been added.
a. 49.99 mL
ml
Solution: The number of moles of H* in 50.00 mL of 1.000 M HCI is
nH. = 50.00 x 10³ L x 1.000 mol H*/1L = 50.00 x 103 mol of H*
in each part of the problem, start 1. By calculating the number of moles of OH added then 2. Find the
number of moles of H* or OH in excess. Finally 3. Calculate [H*] and the pH.
a. (1) noh- = 49.99 x 10³ L x 1.000 mol OH/1L = 49.99 x 10³ mol of OH
(2) because H* and OH react in a 1:1 mole ratio,
NH+ reacted = 49.99 x 103 mol
nH+ excess = 50.00 x 103 mol – 49.99 x 103 mol = 0.01 x 103 = 1 x 105 mol
(3) the total volume is almost exactly 100 mL ( 50.00 mL + 49.99 mL):
[H] = 1 x 105 mol H*/1 x 101 = 1 x 104 M
pH = 4
b. (1) noH- = 50.01 x 10³L x 1.000 mol OH/ 1L = 50.01 x 10³ mol of OH-
(2) nH+ reacted = 50.00 x 10³ mol
noH- excess = 50.01 x 10³ mol - 50.00 x 10³ mol = 0.01 x 10³ mol = 1 x 105 mol
(3) [OH-] = 1 x 10$ mol OH-/1 x 101 L = 1x 104 M
pOH = 4.0; pH = 10.0
Transcribed Image Text:ACID-BASE TITRATIONS
The emphasis is on the equilibrium principles that apply to the acid-base reactions involved.
Strong Acid-Strong Base
The neutralization reaction that takes place when any strong acid reacts with any strong base can be
represented by a net ionic equation of the Bronsted-Lowry type: H*(aq)
OH (aq) → H2O
The equation is the reverse for the ionization of water, so the equilibrium constant can be calculated by
using the reciprocal rule. K= 1/Kw = 1/(1.0 x 10-14 ) = 1.0 x 1014
The enormous value of K means that for all practical purposes this reaction goes to completion,
consuming the limiting reactant, H* or OH .
Consider now what happens when HCI, a typical strong acid, is titrated with NaOH. The figure below
shows how the pH changes during the titration. Two features of this curve are of particular importance:
(1) at the equivalence point, when all the HCI has been neutralized by NaOH, a solution of NaCI, a neutral
salt, is present. The pH at the equivalence point is 7.
(2) near the equivalence point, the pH rises very rapidly. The pH may increase by as much as six units
(from 4 to 10) when half a drop, = 0.02 mL, of NaOH is added.
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