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IgorLugansk [536]
3 years ago
14

Calculate the change in pH when 71.0 mL of a 0.760 M solution of NaOH is added to 1.00 L of a solution that is 1.0O M in sodium

acetate and 1.00 M in acetic acid.
Chemistry
1 answer:
Eddi Din [679]3 years ago
8 0

Explanation:

It is known that pK_{a} value of acetic acid is 4.74. And, relation between pH and pK_{a} is as follows.

                    pH = pK_{a} + log \frac{[CH_{3}COOH]}{[CH_{3}COONa]}

                          = 4.74 + log \frac{1.00}{1.00}

So, number of moles of NaOH = Volume × Molarity

                                                   = 71.0 ml × 0.760 M

                                                    = 0.05396 mol

Also, moles of  CH_{3}COOH = moles of CH_{3}COONa

                                          = Molarity × Volume

                                          = 1.00 M × 1.00 L

                                          = 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

            CH_{3}COONa + NaOH \rightarrow CH_{3}COOH

Initial :    1.00 mol                                  1.00 mol

NaoH addition:               0.05396 mol

Equilibrium : (1 - 0.05396 mol)    0           (1.00 + 0.05396 mol)

                    = 0.94604 mol                       = 1.05396 mol

As, pH = pK_{a} + log \frac{[CH_{3}COONa]}{[CH_{3}COOH]}

               = 4.74 +  log \frac{0.94604}{1.05396}

               = 4.69

Therefore, change in pH will be calculated as follows.

                         pH = 4.74 - 4.69

                               = 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.

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Balance the following Equation:
pashok25 [27]

Answer:

HCl

Explanation:

Given data:

Mass of Zn = 50 g

Mass of HCl = 50 g

Limiting reactant = ?

Solution:

Chemical equation:

Zn + 2HCl      →     ZnCl₂ + H₂

Number of moles of Zn:

Number of moles = mass / molar mass

Number of moles = 50 g/ 65.38 g/mol

Number of moles = 0.76 mol

Number of moles of HCl:

Number of moles = mass / molar mass

Number of moles = 50 g/ 36.5 g/mol

Number of moles = 1.4 mol

Now we will compare the moles of Reactant with product.

                 Zn         :          ZnCl₂

                  1           :             1

                 0.76     :           0.76

                Zn         :             H₂

                  1           :             1

                 0.76     :           0.76

               HCl         :          ZnCl₂

                  2           :             1

                 1.4         :           1/2×1.4 = 0.7

                HCl         :             H₂

                  2           :             1

                 1.4         :           1/2×1.4 = 0.7

Less number of moles of product are formed by HCl it will act limiting reactant.

6 0
3 years ago
An example of a homogeneous mixture is 1. oil and vinegar dressing. 2. concrete. 3. a mixture of salt and pepper. 4. salt water
Sidana [21]

A mixture in which composition is constant throughout the mixture is said to be homogeneous mixture.

Now, oil and vinegar dressing is not considered as a homogeneous mixture because composition is not uniform.

Concrete is made up of three components. Thus, composition is not uniform, so it is not a homogeneous mixture.

A mixture of salt and pepper, when salt is mixed in pepper then the mixture is not uniform as white flecks of salt and black flecks of pepper can be seen in the mixture. Thus, it is not a homogeneous mixture.

Salt water is uniform mixture. Thus,  it is a homogeneous mixture.

Thus, salt water is the correct answer.


4 0
3 years ago
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andrey2020 [161]

Answer:

D bohr

Explanation:

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6 0
3 years ago
Read 2 more answers
A buffer can be prepared by mixing two solutions. Determine if each of the following mixtures will result in a buffer solution o
cestrela7 [59]

Answer:

1) A. "Yes, it will result in a buffer solution."

2) A. "Yes, it will result in a buffer solution."

3) A. "No, it will not result in a buffer solution."

4) B. No, it will not result in a buffer solution."

5) B. No, it will not result in a buffer solution.

Explanation:

A buffer system is formed by two components:

  • A weak acid and its conjugate base.
  • A weak base and its conjugate acid.

<em>1) Mixing 100.0 mL of 0.1 M HF with 100.0 mL of 0.05 M mol KF. </em>A. "Yes, it will result in a buffer solution."

HF is a weak acid and F⁻ is its conjugate base.

<em>2) Mixing 100.0 mL of 0.1 M NH₃ with 100.0 mL of 0.1 M NH₄Br.</em> A. "Yes, it will result in a buffer solution."

NH₃ is a weak base and NH₄⁺ is its conjugate acid.

<em>3) Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.05 M KOH.</em> A. "No, it will not result in a buffer solution."

It does not have the components of a buffer system.

<em>4) Mixing 100.0 mL of 0.1 M HCl with 100.0 mL of 0.1 M KCl.</em> B. No, it will not result in a buffer solution."

It does not have the components of a buffer system (HCl is a strong acid).

<em>5) Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.1 M KOH.</em> B. No, it will not result in a buffer solution.

It does not have the components of a buffer system.

3 0
3 years ago
Joe tries to neutralize 125mL of 2.0 M HCL with 175mL of 1.0M NaOH. will the neutralization reaction be completed? If not what i
Lemur [1.5K]

Answer:

1) The neutralization reaction will mot be completed.

2) pH = 0.6.

Explanation:

<em>1) Will the neutralization reaction be completed?</em>

For the neutralization reaction be completed; The no. of millimoles of the acid must be equal the no. of millimoles of the base.

The no. of millimoles of 125 mL of 2.0 M HCl = MV = (2.0 M)(125.0 mL) = 250.0 mmol.

The no. of millmoles of 175 mL of 1.0 M NaOH = MV = (1.0 M)(175.0 mL) = 175.0 mmol.

∴ HCl will be in excess.

∴ The neutralization reaction will mot be completed.

<em>2) If not what is the pH of the final solution?</em>

[H⁺] =[ (MV)HCl - (MV)NaOH]/V total = (250.0 mmol - 175.0 mmol) / (300.0 mL) = 0.25 M.

∵ pH = - log[H⁺]

<em>∴  pH =</em> - log(0.25) =<em> 0.6.</em>

4 0
3 years ago
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