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Katarina [22]
3 years ago
12

A solution of acetic acid is prepared in water by adding 11.1 g of sodium acetate to 200 ml of water. The final ph is measured t

o be 6.5. What are the concentrations of acetate and acetic acid in solution?
Chemistry
1 answer:
morpeh [17]3 years ago
4 0

Given buffer contains:

Acetic acid: CH3COOH.

Sodium Acetate : CH3COONa.

Mass of CH3COONa taken = 11.1 g

Molar mass of CH3COONa = 82.03 g/mol

# moles of CH3COONa = 11.1g/82.03 g.mol-1 = 0.1353 moles

Volume of water = 200 ml = 0.2 L

Therefore, the initial concentration of CH3COONa = 0.1353/0.2 = 0.677 M

Buffer pH = 6.5

pKa (CH3COOH) = 4.76

Based on Henderson-Hasselbalch equation:

pH = pKa + log [CH3COONa]/[CH3COOH]

Now, initially there is only 0.677 M CH3COONa . If 'x' is the concentration of the protonated species (i.e CH3COOH) formed, then the amount of CH3cooNa remaining = 0.677-x

6.5 = 4.76 + log [0.677-x]/[x]

x = 0.011 M

i.e. [CH3COOH] = 0.011 m

[CH3COONa] = 0.677-0.011 = 0.666 M

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Answer:

Ka=3.98x10^{-4}

Explanation:

Hello there!

In this case, since the modelling of titration problems can be approached via the Henderson-Hasselbach equation to set up a relationship between pH, pKa and the concentration of the acid and its conjugate base, we can write:

pH=pKa+log(\frac{[NO_2^-]}{[HNO_2]} )

Whereas the pH is given as 3.14 and the concentrations are the same, that is why the pH would be equal to the pKa as the logarithm gets 0 (log(1)=0); thus, we can calculate the Ka via:

Ka=10^{-pKa}=10^{-3.14}\\\\Ka=3.98x10^{-4}

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Copper has two naturally occurring isotopes, 63Cu Gsotopic mass 629396 arnu) and 65Cu Osotopic mass 64.9278 amu). If copper has
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<u>Answer:</u> The percentage abundance of _{29}^{63}\textrm{Cu} and _{29}^{65}\textrm{Cu} isotopes are 75.77% and 24.23% respectively

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of _{29}^{63}\textrm{Cu} isotope be 'x'. So, fractional abundance of _{29}^{63}\textrm{Cu} isotope will be '1 - x'

  • <u>For _{29}^{63}\textrm{Cu} isotope:</u>

Mass of _{29}^{63}\textrm{Cu} isotope = 62.9396 amu

Fractional abundance of _{29}^{63}\textrm{Cu} isotope = x

  • <u>For _{29}^{65}\textrm{Cu} isotope:</u>

Mass of _{29}^{65}\textrm{Cu} isotope = 64.9278 amu

Fractional abundance of _{29}^{65}\textrm{Cu} isotope = 1 - x

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Putting values in equation 1, we get:

63.546=[(62.9396\times x)+(64.9278\times (1-x))]\\\\x=0.6950

Percentage abundance of _{29}^{63}\textrm{Cu} isotope = 0.6950\times 100=69.50\%

Percentage abundance of _{29}^{65}\textrm{Cu} isotope = (1-0.6950)=0.305\times 100=30.50\%

Hence, the percentage abundance of _{29}^{63}\textrm{Cu} and _{29}^{65}\textrm{Cu} isotopes are 69.50% and 30.50% respectively.

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