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soldi70 [24.7K]
2 years ago
13

A formic acid buffer solution contains 0. 20 m h c o o h hcooh and 0. 24 m h c o o − hcoox−. the pka of formic acid is 3. 75. wh

at is the ph of the buffer?
Chemistry
1 answer:
irina1246 [14]2 years ago
6 0

A buffer solution contains an equivalent amount of acid and base. The pH of the solution with an acid dissociation constant (pKa) value of 3.75 is 3.82.

<h3>What is pH?</h3>

The amount of hydrogen or the proton ion in the solution is expressed by the pH. It is given by the sum of pKa and the log of the concentration of acid and bases.

Given,

Concentration of salt [HCOO⁻] = 0.24 M

Concentration of acid [HCOOH] = 0.20 M

The acid dissociation constant (pKa) = 3.75

pH is calculated from the Hendersons equation as,

pH = pKa + log [salt] ÷ [acid]

pH = 3.75 + log [0.24] ÷ [0.20]

= 3.75 + log (1.2)

= 3.75 + 0.079

= 3.82

Therefore, 3.82 is the pH of the buffer.

Learn more about pH here:

brainly.com/question/27181245

#SPJ4

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How can you tell from the name the types of bonds present in a hydrocarbon?
Aneli [31]

Answer:

Of bond is formed by sharing of electrons - covalent

If bond is formed by donation and by accepting electrons - ionic bond

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Which element is in the same family as Chlorine (Cl) and Fluorine (F)?
ValentinkaMS [17]

Answer/Explanation:

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Bromine (Br)

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This is due at 12am HELP!!
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4 0
3 years ago
An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?
umka2103 [35]

<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

K_b = molal boiling point elevation constant = 0.51°C/m

m_{solute} = Given mass of solute (CsCl) = 8.00 g

M_{solute} = Molar mass of solute (CsCl) = 168.4  g/mol

W_{solvent} = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC

Hence, the boiling point of solution is 100.53

6 0
3 years ago
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