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damaskus [11]
3 years ago
8

A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use 1.80 x 10−5 as Ka for ac

etic acid. (a) What is the pH of the solution? (b) Is the solution acidic or basic? (c) What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to 0.200 L of the original buffer
Chemistry
1 answer:
Anika [276]3 years ago
8 0

Answer:

a. pH = 5.22

b. Acidic.

c. pH = 5.14

Explanation:

a. It is possible to find the pH of a buffer using Henderson-Hasselbalch equation (H-H equation):

pH = pKa + log₁₀ [A⁻] / [HA]

<em>Where pKa is -log Ka (For acetic acid =  4.74), [A⁻] is molar concentration of conjugate base (Acetate salt) and [HA] concentration of the weak acid (Acetic acid).</em>

Replacing:

pH = 4.74 + log₁₀ [0.600M] / [0.200M]

<em>You use the concentration of the acetic acid and sodium acetate because you're adding equal volumes, doing the ratio of the species the same</em>

<em />

<h3>pH = 5.22</h3><h3 />

b. As the solution has a pH lower that 7.0, it is considered as a <em>acidic solution.</em>

<em></em>

c. When you add HCl to the buffer, the reaction is:

CH₃COO⁻ + HCl → CH₃COOH + Cl⁻

<em>Where acetate ion reacts with the acid producing acetic acid.</em>

As you have 0.200L of the buffer, 0.100L are of the acetate ion and 0.100L of the acetic acid. Initial moles of both compounds and moles of HCl added are:

CH₃COO⁻: 0.100L ₓ (0.600mol / L) = 0.0600 moles

CH₃COOH: 0.100L ₓ (0.200mol / L) = 0.0200 moles

HCl: 3.0mL = 3x10⁻³L ₓ (0.034mol / L) =  0.00010 moles HCl

The moles added of HCl are the same moles you're consuming of acetate ion and producing of acetic acid. Thus, moles after the reaction are:

CH₃COO⁻: 0.0600 moles - 0.0001 moles = 0.0509 moles

CH₃COOH: 0.0200 moles + 0.0001 moles = 0.0201 moles

Replacing in H-H equation:

pH = 4.74 + log₁₀ [0.0509moles] / [0.0201moles]

<h3>pH = 5.14</h3>

<em />

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3 years ago
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans
Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

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0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

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We are given:

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4 0
3 years ago
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frutty [35]

Answer:

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7 0
3 years ago
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Hunter-Best [27]

Answer:

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The number nitrogen atoms are six. Elena did mistake by counting the number of nitrogen. She should didn't multiplied the nitrogen atom by subscript 2.

5 0
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aalyn [17]
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carbon was oxidized
5 0
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