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
Evgen [1.6K]
3 years ago
8

What is the pH of a solution prepared by mixing 25.00 mL of 0.10 M CH3CO2H with 25.00 mL of 0.010 M CH3CO2Na? Assume that the vo

lume of the solutions are additive and that K a = 1.8 × 10-5 for CH3CO2H.
Chemistry
1 answer:
denpristay [2]3 years ago
4 0

Answer:

pH = 3.74

Explanation:

<u>Given:</u>

Initial volume of CH3COOH, V1 = 25.00 ml

Initial concentration of CH3COOH, M1 = 0.10 M

Initial volume of CH3COONa, V1 = 25.00 ml

Initial concentration of CH3COONa, M2 = 0.010 M

Ka (CH3COOH) = 1.8*10^-5

<u>To determine:</u>

pH of the solution

<u>Calculation:</u>

The given solution of CH3COOH/CH3COONa is in fact a buffer whose pH is given by the Henderson-Hasselbalch equation where:

pH = pKa + log\frac{[A-]}{[HA]} ----(1)

where A- = concentration of conjugate base = [CH3COONa]

HA = weak acid = [CH3COOH]

Step 1: Calculate the final concentration of CH3COONa

V1 = 25.00 ml

V(final) = Total volume = 25.00 + 25.00 = 50.00 ml

M1 = 0.010 M

M1V1 = M2V2\\\\M2 = \frac{M1V1}{V2} = \frac{0.010 M * 25.00 ml}{50.00ml} =0.005M

<u>Step 2</u>: Calculate the final concentration of CH3COOH

V1 = 25.00 ml

V(final) = Total volume = 25.00 + 25.00 = 50.00 ml

M1 = 0.10 M

M1V1 = M2V2\\\\M2 = \frac{M1V1}{V2} = \frac{0.10 M * 25.00 ml}{50.00ml} =0.05M

<u>Step 3</u>: Calculate the pH

Based on equation (1)

pH = pKa + log\frac{[CH3COONa]}{[CH3COOH]} ----(1)

pKa = -log Ka = -log(1.8*10^-5) = 4.74

pH = -logKa + log\frac{[CH3COONa]}{[CH3COOH]}

pH = 4.74 + log\frac{[0.005]}{[0.05]}

pH = 3.74

You might be interested in
Solution A has twice as much solute as solution B is it possible for the solutions to have the same concentration
pishuonlain [190]
Yes. As long as the ratio of solute and solvent is same for both solution, the solution has the same concentration. for example, solution A has 2 ml of water, and 1 ml of sucrose. Solution B has 4ml of water and 2ml of sucrose. Both has a ratio of water to sucrose by 2 : 1. they have the same concentration of 50% sucrose. 
5 0
3 years ago
HELP PLEASE I NEED HELP THANKS I LOVE U
Pani-rosa [81]

Answer:

0.500-Molarity solution

Explanation:

4 0
3 years ago
Given the unbalanced equation: ___Al2(SO4)3+___Ca(OH)2—&gt;___Al(OH)3+__CaSO4
nasty-shy [4]

Answer:

The answer to your question is letter B. 9

Explanation:

Unbalanced reaction

                     Al₂(SO₄)₃  +  Ca(OH)₂   ⇒    Al(OH)₃   +   CaSO₄

                     Reactants             Elements      Products

                          2                          Al                     1

                          3                           S                     1

                         14                           O                    7

                          1                           Ca                    1

                          2                           H                     3

Balanced reaction

                    Al₂(SO₄)₃  + 3Ca(OH)₂   ⇒   2Al(OH)₃   +   3CaSO₄

                     Reactants             Elements      Products

                          2                          Al                    2

                          3                           S                    3

                         18                           O                  18

                          3                           Ca                  3

                          6                            H                    6

The sum of the coefficients is 1 + 3+ 2+ 3 = 9

8 0
3 years ago
The Charles Zeppelin, a fictional airship, is filled with 2.16 x 105 liters of hydrogen gas (H2). On the ground, the airship’s t
Dmitry [639]

Answer:

V2 = 3.11 x 105 liters

Explanation:

Initial Volume, V1 = 2.16 x 105 liters

Initial Temperature, T1 = 295 K

Final Temperature, T2 = 425 K

Final Volume, V2 = ?

These quantities are related by charle's law and the equation of the law is given as;

V1 / T1  =  V2 / T2

V2 = T2 * V1 / T1

V2 = 425 * 2.16 x 105 / 295

V2 = 3.11 x 105 liters

8 0
3 years ago
Fermeroben heres the messages COUGH
lord [1]

I completely understand your gripe. I also believe that you are the better option. It will only take time until she is yours, despite your parent's views. Keep it up

6 0
3 years ago
Other questions:
  • Sea water has about 56.0 grams of NaCl for every 2.0 liters of water. what is the molarity?
    12·1 answer
  • A solution is prepared by adding 0.10 mole of Ni(NH3)6Cl2 to 0.50 L of 3.0 M NH3. Calculate [Ni(NH3)62 ] and [Ni2 ] in this solu
    14·1 answer
  • Why is water considered a polar molecule?
    6·1 answer
  • What is needed to change a phase of matter?
    6·1 answer
  • Water can keep engines operating at low temperatures. Which property of water most likely gives it this ability?
    11·1 answer
  • Which of the following substances feels slippery and will not react with metals? *
    13·1 answer
  • A student is building a simple circuit with a battery, light bulb, and copper wires. When she connects the wires to the battery
    10·1 answer
  • An electrically neutral atom consists of 15 neutrons, 13 electrons, and a number
    6·1 answer
  • How many particles are in 2.4 moles of carbon dioxide
    5·1 answer
  • If 5 mol of oxygen gas effuses through an opening in 10 seconds, how long will it take for the same amount of hydrogen gas to ef
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!