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SVEN [57.7K]
3 years ago
11

an acid based titration was performed. it took 27.45 mL of the base, KOH, to titrate 3.115 g HBr, the acid. what was the molarit

y of KOH?​
Chemistry
1 answer:
sveticcg [70]3 years ago
7 0

Answer:

<u>1.4 M</u>

Explanation:

n(HBr)=3.115/81

so, 3.115/81=0.0385mol

according to the reaction, n(HBr)=n(KOH)=0.038 mol

C(KOH)=n/V=0.0385/0.02745

0.0385/0.02745 =1.4 M

in short the answer is 1.4 M (molarity)

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1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium
lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

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3 years ago
Define an independent variable in an experiment. Explain the term with respect to the solubility of salt at different temperatur
nekit [7.7K]

Independent variable would be salt since you can't change it in this experiment.

6 0
3 years ago
When electricity flows through water, the water molecules split into molecules of hydrogen and oxygen gas. How many oxygen (O2)
ikadub [295]
That would be one oxygen molecule.
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3 years ago
Read 2 more answers
Need help setting the problem up
Sveta_85 [38]

Answer:

4.0 moles

Explanation:

The following data were obtained from the question:

Volume (V) = 12L

Pressure = 5.6 atm

Temperature (T) = 205K

Gas constant (R) = 0.08206 atm.L/Kmol

Number of mole (n) =?

Using the ideal gas equation: PV = nRT, the number of mole of the gas can be obtained as follow

PV = nRT

5.6 x 12 = n x 0.08206 x 205

Divide both side by 0.08206 x 205

n = (5.6 x 12)/(0.08206 x 205)

n = 4.0 moles

Therefore, the number of mole of the gas is 4.0 moles

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3 years ago
Which are examples of contact forces?
lesantik [10]

Answer:

Contact force is a force that requires a contact between two bodies and is ubiquitous in nature

Explanation:

The force is distributed in two categories i.e

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Contact Forces  are the forces which is requires a contact between two bodies to occur.

The Types of contact forces are given below:

  • Frictional Force
  • Tension Force
  • Normal Force
  • Air Resistance Force
  • Applied Force
  •  Spring Force

Some common daily examples of the contact force are

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  2. Air flowing in opposite direction of a moving object.
  3. Pushing a table with hand and the friction between its base and floor.
  4. Stretching a rubber band with hands etc.

Also non-contact forces are Gravitational force, Electrical force and Magnetic force.

8 0
3 years ago
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