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Phantasy [73]
3 years ago
14

Calculate the pH of a solution prepared by dissolving 0.370 mol of formic acid (HCO2H) and 0.230 mol of sodium formate (NaCO2H)

in water sufficient to yield 1.00 L of solution. The Ka of formic acid is
Chemistry
1 answer:
Veronika [31]3 years ago
5 0

The question is incomplete, here is the complete question:

Calculate the pH of a solution prepared by dissolving 0.370 mol of formic acid (HCO₂H) and 0.230 mol of sodium formate (NaCO₂H) in water sufficient to yield 1.00 L of solution. The Ka of formic acid is 1.77 × 10⁻⁴

a) 2.099

b) 10.463

c) 3.546

d) 2.307

e) 3.952

<u>Answer:</u> The pH of the solution is 3.546

<u>Explanation:</u>

We are given:

Moles of formic acid = 0.370 moles

Moles of sodium formate = 0.230 moles

Volume of solution = 1 L

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

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

pH=pK_a+\log(\frac{[salt]}{[acid]})

pH=pK_a+\log(\frac{[HCOONa]}{[HCOOH]})

pK_a = negative logarithm of acid dissociation constant of formic acid = 3.75

[HCOONa]=\frac{0.230}{1}  

[HCOOH]=\frac{0.370}{1}

pH = ?  

Putting values in above equation, we get:  

pH=3.75+\log(\frac{0.23/1}{0.37/1})\\\\pH=3.54

Hence, the pH of the solution is 3.546

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3 years ago
Part 1: A cylinder containing 20.0 L of compressed nitrogen is connected to an empty (evacuated) vessel with an unknown volume.
baherus [9]

Answer:

The volume of the vessel is 250 L

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Explanation:

Using Boyle's law  

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P₂ = 2 atm

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{25}\times {20.0}={2}\times {V_2}

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<u>The volume of the vessel is 250 L.</u>

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P_{H_2}=Mole\ fraction\times Total\ Pressure

So, according to definition of mole fraction:

Mole\ fraction\ of\ H_2=\frac {n_{H_2}}{n_{H_2}+n_{He}}

Also,

Mole fraction of H₂ = 1 - Mole fraction of He = 1 - 0.75 = 0.25

So,

Total pressure = 756 torr

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3 years ago
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svlad2 [7]

Answer:

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Explanation:

As concentrations of two aqueous solutions are same therefore we can write:

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Here both Na_{2}CO_{3} and LiCl are strong electrolytes.

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Answer:

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