1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
aniked [119]
2 years ago
15

A 25.0 ml sample of 0.150 m benzoic acid is titrated with a 0.150 m naoh solution. what is the ph at the equivalence point? the

ka of benzoic acid is 4.50 × 10-4.
Chemistry
2 answers:
olya-2409 [2.1K]2 years ago
8 0

Answer:

pH = 8.11

Explanation:

Step 1: Data given

Volume of a 0.150 M benzoic acid sample = 25.0 mL

Molarity of NaOH solution = 0.150 M

Ka of the benzoic acid = 4.50 * 10^-4

Step 2: The balanced equation

C7H6O2 + NaOH → C7H5O2Na  + H2O

Step 3: calculate moles

At the equivalence point we have the same amount of of acid and base

Both benzoic acid and NaOH will completely be consumed.

Moles benzoic acid = molarity *volume

Moles benzoic acid = 0.150 M * 0.025 L

Moles benzoic acid = 0.00375 moles

We we'll need 0.00375 moles NaOH to reach the equivalence point

Volume NaOH = 0.00375 moles / 0.150 M

Volume = 0.025 L = 25 mL

Total volume = 25 + 25 = 50 mL

Step 4: Calculate moles of C7H5O2Na

For 1 mol benzoic acid we need 1 mol NaOH to produce 1 mol C7H5O2Na

For 0.00375 moles benzoic acid we'll have 0.00375 moles C7H5O2Na

Step 5: The equation

C7H5O2- + H2O → C7H6O2 + OH-

Step 6: The initial concentration

[C7H5O2-] = 0.00375 moles / 0.050 L

[C7H5O2-] = 0.075 M

[C7H6O2] = 0M

[OH-] = 0M

Step 7: The concentration at the equilibrium

[C7H5O2-] = 0.075 - X M

[C7H6O2] = XM

[OH-] = XM

Step 8: Calculate Kb

Ka * Kb = Kw

Kb = Kw / Ka

Kb = 10^-14 / (4.50 *10^-4)

Kb = 2.22 *10^-11

Kb =  [C7H6O2][OH-]/[C7H5O2-]

2.22 * 10^-11 = X²/ 0.075 - X

2.22 * 10^-11 = X²/ 0.075

X² = 2.22 *10^-11 * 0.075

X² = 1.665 *10^-12

X = 1.29 * 10^-6 = [OH-]

Step 9: Calculate pOH

pOH = -log[OH-]

pOH = -log(1.29 * 10^-6 )

pOH = 5.89

Step 10: Calculate pH

pH = 14 - 5.89

pH = 8.11

Ad libitum [116K]2 years ago
5 0

Solution:

At the equivalence point, moles NaOH = moles benzoic acid  

HA + NaOH ==> NaA + H2O where HA is benzoic acid  

At the equivalence point, all the benzoic acid ==> sodium benzoate  

A^- + H2O ==> HA + OH- (again, A^- is the benzoate anion and HA is the weak acid benzoic acid)  

Kb for benzoate = 1x10^-14/4.5x10^-4 = 2.22x10^-11  

Kb = 2.22x10^-11 = [HA][OH-][A^-] = (x)(x)/0.150  

x^2 = 3.33x10^-12  

x = 1.8x10^-6 = [OH-]  

pOH = -log [OH-] = 5.74  

pH = 14 - pOH = 8.26  


You might be interested in
What is the equilibrium constant of a reaction?
xeze [42]

Answer:

A. It is the ratio of the concentrations of products to the concentrations of reactants.

Explanation:

The equilibrium constant of a chemical reaction is the ratio of the concentration of products to the concentration of reactants.

This equilibrium constant can be expressed in many different formats.

  • For any system, the molar concentration of all the species on the right side are related to the molar concentrations of those on the left side by the equilibrium constant.
  • The equilibrium constant is a constant at a given temperature and it is temperature dependent.
  • The derivation of the equilibrium constant is based on the law of mass action.
  • It states that "the rate of a chemical reaction is proportional to the product of the concentration of the reacting substances. "
8 0
3 years ago
Distilled water density 1.0 g/cm^3 propane density 0.494 g/cm^3 salt water density 1.025 g/cm^3 liquid gold density 17.31g/cm^3
I am Lyosha [343]
What are you asking?
6 0
3 years ago
You need 270 ml of a 65% alcohol solution. on hand, you have a 90% alcohol mixture. how much of the 90% alcohol mixture and pure
olganol [36]

You must add 75 mL water to 195 mL 90 % alcohol to make 270 mL of 65 % alcohol.

<em>Step 1.</em> Calculate the volume of 90 % alcohol needed

You can use the dilution formula

<em>V</em>1×<em>C</em>1 = <em>V</em>2×<em>C</em>2

where

<em>V</em>1 and<em> V</em>2 are the volumes of the two solutions

<em>C</em>1 and <em>C</em>2 are the concentrations

You can solve the above formula to get

<em>V</em>2 = <em>V</em>1 × <em>C</em>1/<em>C</em>2

<em>V</em>1 = 270 mL; <em>C</em>1 = 65 %

V2 = ?; _____<em>C</em>2 = 90 %

∴<em>V</em>2 = 270 mL × (65 %/90 %) = 195 mL

You need 195 mL of 90 % alcohol to make 270 mL of 65 % RA

<em>Step 2</em>. Calculate the amount of water to add.

Volume of water = 270 mL – 195 mL = 75 mL

8 0
3 years ago
A baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby weighs 12 N. The carriage has ____
QveST [7]

Answer:

B

Explanation:

The carriage has potential energy

5 0
2 years ago
Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re
borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

                        A            +            B           ⇄        C

where;

numbers of moles = 0.386 mol C  (g)

Volume =  7.29 L

Molar concentration of C = \frac{0.386}{7.29}

= 0.053 M

                        A            +            B           ⇄        C

Initial               0                           0                      0.530    

Change          +x                          +x                       - x

Equilibrium      x                           x                      (0.0530 - x)

K = \frac{[C]}{[A][B]}

where

K is given as ; 78.2 atm-1.

So, we have:

78.2=\frac{[0.0530-x]}{[x][x]}

78.2= \frac{(0.0530-x)}{(x^2)}

78.2x^2= 0.0530-x

78.2x^2+x-0.0530=0  

Using quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

where; a = 78.2 ; b = 1 ; c= - 0.0530

= \frac{-b+\sqrt{b^2-4ac} }{2a}   or \frac{-b-\sqrt{b^2-4ac} }{2a}

= \frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}  or \frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}

= 0.0204  or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

     = 0.0530 - 0.0204

     = 0.0326

Total number of moles at equilibrium = 0.0204 +  0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

V = \frac{nRT}{P}

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K  (fixed constant temperature )

V (volume) = ???

V=\frac{(0.0734*0.0821*273.15)}{(1.00)}

V = 1.64604

V ≅ 1.65 L

3 0
3 years ago
Other questions:
  • What results electrons into having light produced by neon gas?
    12·1 answer
  • What do we mean by atmospheric pressure
    15·1 answer
  • Early Earth was largely anoxic with very little oxygen present in the atmosphere. However, approximately 2 billion years ago, a
    14·1 answer
  • I am something that is alive or was once alive.
    7·2 answers
  • They dump large amounts of rainwater that causes flooding. *
    10·2 answers
  • Based on your observations of the laboratory assignment(s) that produced gases, can you conclude
    14·1 answer
  • List at least two chemical reactions that resulted in the yellow orange or red preList at least two chemical reactions that resu
    12·1 answer
  • Object A and B have the same mass. Which object is experiencing a stronger gravitational force?
    9·1 answer
  • Explain why in the early morning just before dawn breaks, the temperature of the air
    11·1 answer
  • 1 loop in the primary coil and 8 loops in the secondary. If the secondary voltage is 120 V, what must be the primary voltage
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!