Answer:
pH = 12.43
Explanation:
...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it.
To solve this question we need to know that hidrazoic acid reacts with KOH as follows:
HN3 + KOH → KN3 + H2O
Moles KOH:
0.5716L * (0.2900mol /L) =0.1658 moles of KOH
Moles HN3:
0.2127L * (0.6800mol/L) = 0.1446 moles HN3
As the reaction is 1:1, the KOH is in excess. The moles in excess of KOH are:
0.1658 moles - 0.1446 moles =
0.0212 mol KOH
In 212.7mL + 571.6mL = 784.3mL = 0.7843L
The molarity of KOH = [OH-] is:
0.0212 mol KOH / 0.7843L = 0.027M = [OH-]
The pOH is defined as -log [OH-]
pOH = -log 0.027M
pOH = 1.57
pH = 14 - pOH
pH = 12.43
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 6.90kg of water at 34.7 degrees C . During the reaction 57.1kJ of heat flows out of the bath and into the flask.
Calculate the new temperature of the water bath. You can assume the specific heat capacity of water under these conditions is 4.18J.g^(-1).K^(-1) . Round your answer to significant digits.
Answer:
[tex]T_2= 36.7 \textdegree C[/tex]
Explanation:
Mass of Water [tex]m_w=6.90kg[/tex]
Temperature [tex]T=34.7 degrees[/tex]
Heat Flow [tex]H=57.1kJ[/tex]
Specific heat capacity of water [tex]\mu= 4.18J.g^(-1).K^(-1)[/tex]
Generally the equation for Final Temperature is mathematically given by
[tex]M*\mu *T_1 + Q = M*\mu *T_2[/tex]
[tex]T_2=\frac{M*\mu *T_1 + Q }{M*\mu}[/tex]
Therefore
[tex]T_2=\frac{6.90*4.18*34.7 + 57.1}{6.90*4.18}[/tex]
[tex]T_2= 36.7 \textdegree C[/tex]
Could someone please help me out???
Answer:
Time is 2.2 seconds.
Explanation:
Time:
[tex]{ \boxed{ \bf{time = \frac{distance}{speed} }}}[/tex]
Substitute into the formula:
speed = 715 km/h = 198.61 m/s
[tex]{ \tt{time = \frac{435}{198.61} }} \\ { \tt{time = 2.2 \: seconds}}[/tex]
A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.60 mL of a 0.400 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
A beaker with 120mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M. A student adds 6.60mL of a 0.300M HCl solution to the beaker. How much will the pH change?
The pKa of acetic acid is 4.76.
Chemistry Buffer Calculations
1 Answer
Stefan V.
May 8, 2016
Δ
pH
=
0.29
Explanation:
!! LONG ANSWER !!
The idea here is that you need to use the Henderson-Hasselbalch equation to determine the ratio that exists between the concentration of the weak acid and of its conjugate base in the buffer solution.
Once you know that, you can use the total molarity of the acid and of the conjugate base to find the number of moles of these two chemical species present in the buffer.
So, the Henderson-Hasselbalch equation looks like this
∣
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
pH
=
p
K
a
+
log
(
[
conjugate base
]
[
weak acid
]
)
a
a
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
In your case, you have acetic acid,
CH
3
COOH
, as the weak acid and the acetate anion,
CH
3
COO
−
, as its conjugate base. The
p
K
a
of the acid is said to be equal to
4.76
, which means that you have
pH
=
4.76
+
log
(
[
CH
3
COO
−
]
[
CH
3
COOH
]
)
The pH is equal to
5
, and so
5.00
=
4.76
+
log
(
[
CH
3
COO
−
]
[
CH
3
COOH
]
)
log
(
[
CH
3
COO
−
]
[
CH
3
COOH
]
)
=
0.24
This will be equivalent to
10
log
(
[
CH
3
COO
−
]
[
CH
3
COOH
]
)
=
10
0.24
which will give you
[
CH
3
COO
−
]
[
CH
3
COOH
]
=
1.74
This means that your buffer contains
1.74
times more conjugate base than weak acid
[
CH
3
COO
−
]
=
1.74
×
[
CH
3
COOH
]
Now, because both chemical species share the same volume,
120 mL
, this can be rewritten as
n
C
H
3
C
O
O
−
120
⋅
10
−
3
L
=
1.74
×
n
C
H
3
C
O
O
H
120
⋅
10
−
3
L
which is
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
n
C
H
3
C
O
O
−
=
1.74
×
n
C
H
3
C
O
O
H
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
(
1
)
So, the buffer contains
1.74
times more moles of acetate anions that of acetic acid.
Now, the total molarity of the buffer is said to be equal to
0.1 M
. You thus have
[
CH
3
COOH
]
+
[
CH
3
COO
−
]
=
0.10 M
Once again, use the volume of the buffer to write
n
C
H
3
C
O
O
H
120
⋅
10
−
3
L
+
n
C
H
3
C
O
O
−
120
⋅
10
−
3
L
=
0.1
moles
L
This will be equivalent to
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
n
C
H
3
C
O
O
−
+
n
C
H
3
C
O
O
H
=
0.012
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
(
2
)
Use equations
(
1
)
and
(
2
)
to find how many moles of acetate ions you have in the buffer
1.74
⋅
n
C
H
3
C
O
O
H
+
n
C
H
3
C
O
O
H
=
0.012
n
C
H
3
C
O
O
H
=
0.012
1.74
+
1
=
0.004380 moles CH
3
COOH
This means that you have
n
C
H
3
C
O
O
−
=
1.74
⋅
0.004380 moles
n
C
H
3
C
O
O
−
=
0.007621 moles CH
3
COO
−
Now, hydrochloric acid,
HCl
, will react with the acetate anions to form acetic acid and chloride anions,
Cl
−
H
Cl
(
a
q
)
+
CH
3
COO
−
(
a
q
)
→
CH
3
COO
H
(
a
q
)
+
Cl
−
(
a
q
)
Notice that the reaction consumes hydrochloric acid and acetate ions in a
1
:
1
mole ratio, and produces acetic acid in a
1
:
1
mole ratio.
Use the molarity and volume of the hydrochloric acid solution to determine how many moles of strong acid you have
∣
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
c
=
n
solute
V
solution
⇒
n
solute
=
c
⋅
V
solution
a
a
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
In your case, this gets you
n
H
C
l
=
0.300 mol
L
−
1
⋅
volume in liters
6.60
⋅
10
−
3
L
n
H
C
l
=
0.001980 moles HCl
The hydrochloric acid will be completely consumed by the reaction, and the resulting solution will contain
n
H
C
l
=
0 moles
→
completely consumed
n
C
H
3
C
O
O
−
=
0.007621 moles
−
0.001980 moles
=
0.005641 moles CH
3
COO
−
n
C
H
3
C
O
O
H
=
0.004380 moles
+
0.001980 moles
=
0.006360 moles CH
3
COOH
The total volume of the solution will now be
V
total
=
120 mL
+
6.60 mL
=
126.6 mL
The concentrations of acetic acid and acetate ions will be
[
CH
3
COOH
]
=
0.006360 moles
126.6
⋅
10
−
3
L
=
0.05024 M
[
CH
3
COO
−
]
=
0.005641 moles
126.6
⋅
10
−
3
L
=
0.04456 M
Use the Henderson-Hasselbalch equation to find the new pH of the solution
pH
=
4.76
+
log
(
0.04456
M
0.05024
M
)
pH
=
4.71
Therefore, the pH of the solution decreased by
Δ
pH
=
|
4.71
−
5.00
|
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
0.29 units
a
a
∣
∣
−−−−−−−−−−−−−
Answer link
Related topic
Buffer Calculations
Questions
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7.7 cm
9.8 cm
0.00
0.162 m
Answer:
Volume = 1222.5cm³
Explanation:
If the question is about the volume of the rectangle:
The volume of a rectangle is obtained by the multiplication of its 3 dimensions: Length, width, height.
In the problem, the length of the rectangle is 0.162m = 16.2cm
The width is 7.7cm
And the height is 9.8cm
The volume is:
Volume = 16.2cm*7.7cm*9.8cm
Volume = 1222.5cm³The reversible reaction 2H2 CO <------> CH3OH heat is carried out by mixing carbon monoxide and hydrogen gases is a closed vessel under high pressure with a suitable catalyst . After equilibrium is established at high temperature and pressure, all three substances are present. If the pressure on the system is lower, with the temperature kept constant, what will be the result
Answer:
The amount of CH3OH present in the mixture would decrease
Explanation:
According to Le Cha-telier's principle, when a reaction is in equilibrium and one of the constraints that influence the rate of reactions is applied, the equilibrium would shift so as to neutralize the effects of the constraint.
In this case, looking at the equation of the reaction:
2H2 + CO <------> CH3OH + heat
the total number of moles on the reactant's (left hand) side is 3 (2+1) while on the product's (right hand) side, it is 1. If the pressure on the system is increased, more CH3OH (and less of H2 and CO) will be produced because its side has the lower number of moles out of the two sides.
If the pressure on the system is otherwise lowered, more of H2 and CO would be produced while the amount of CH3OH present would gradually decrease.
Which of the following will affect the rate of a chemical reaction?
solution temperature
solution color
solute mass
solution volume
Answer:
Solution temperature.
Explanation:
Hello there!
In this case, considering this question about chemical kinetics, it will be possible for us to analyze two perspectives:
1. Molecular: here, we infer that the solution temperature will provide more energy to the molecules in order to undergo more effective crashes which will make more products and therefore, increase the rate constant.
2. Mathematical: by means of the Arrhenius equation, it will be possible to tell that the increase in the temperature of the system, the negative of the exponent present in such equation will increase and therefore turn the rate constant bigger.
In such way, we infer the answer is solution temperature.
Regards!
