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defon
3 years ago
8

A buffer solution is composed of 4.00 4.00 mol of acid and 3.25 3.25 mol of the conjugate base. If the p K a pKa of the acid is

4.70 4.70 , what is the pH of the buffer?
Chemistry
1 answer:
Reika [66]3 years ago
8 0

<u>Answer:</u> The pH of the buffer is 4.61

<u>Explanation:</u>

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{conjuagate base}]}{[\text{acid}]})

We are given:

pK_a = negative logarithm of acid dissociation constant of weak acid = 4.70

[\text{conjuagate base}]} = moles of conjugate base = 3.25 moles

[\text{acid}]  = Moles of acid = 4.00 moles

pH = ?

Putting values in above equation, we get:

pH=4.70+\log(\frac{3.25}{4.00})\\\\pH=4.61

Hence, the pH of the buffer is 4.61

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6.25 L

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

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2 years ago
How many protons, neutrons, and electrons are in a neutral atom with the symbol 41Ca
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- Explain how ionic bonds, covalent bonds, and metallic bonds each form and how.
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-Covalent Bonds.

Covalent Compounds. Contain no metals and no ions. Covalent compounds contain nonmetals only.

Example:

Ionic Compounds. A metal with a non-metal. Doesn't use prefixes for naming. Name the metal and change the nonmetal ending to -ide.

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6 0
2 years ago
How many molecules are in 7.62 L of CH4, at 87.5°C and 722 torr
pickupchik [31]

Answer: There are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

Explanation:

Given : Volume = 7.62 L

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Pressure = 722 torr

1 torr = 0.00131579

Converting torr into atm as follows.

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where,

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T = temperature

Substitute the values into above formula as follows.

PV = nRT\\0.95 atm \times 7.62 L = n \times 0.0821 L atm/mol K \times 360.5 K\\n = \frac{0.95 atm \times 7.62 L}{0.0821 L atm/mol K \times 360.5 K}\\= \frac{7.239}{29.59705}\\= 0.244 mol

According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms. Hence, number of atoms or molecules present in 0.244 mol are calculated as follows.

0.244 mol \times 6.022 \times 10^{23}\\= 1.469 \times 10^{23}

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5 0
2 years ago
Given the following UNBALANCED reaction: NH3 (g) &lt;--&gt; N2 (g) + H2 (g) If 1
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Answer:

C. 1.35

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                                                     2NH3 (g) <-->          N2 (g) +             3H2 (g)

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change in concentration        2x                             x                           3x

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concentration

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