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3 years ago

Example of false solutions and true solutions

1 answer:

5 0

False solutions are natural gas, nuclear power ‎ and industrial biomass incineration.

sugar in water is an example of true solution.

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An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni

Svetllana [295]

Answer: The pH of acid solution is 4.58

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KOH:

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

$1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol$

For propanoic acid:

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

$0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol$

The chemical reaction for propanoic acid and KOH follows the equation:

$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

Initial: 0.1372 0.04514

Final: 0.09206 - 0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L (Conversion factor: 1 L = 1000 mL)

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

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

$pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

$[C_2H_5COOK]=\frac{0.04514}{0.26594}$

$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

pH = ?

Putting values in above equation, we get:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

7 0

3 years ago

Which of the following is best revealed by the bohr model?

N76 [4]

D. Electrons orbit in a fixed shell

4 0

3 years ago

Read 2 more answers

Select all that apply.

kap26 [50]

I think it is A and C!

3 0

2 years ago

What is the mass, in grams, of a sample of 1.20 × 1022 atoms of mercury (Hg)? Show your work or explain the steps that you used

o-na [289]

Atomic mass Hg = 200.59 u.m.a

200.59 g --------------- 6.02x10²³ atoms
( mass Hg ) ----------- 1.20 x10²² atoms

mass Hg = ( 1.20x10²² ) x 200.59 / 6.02x10²³

mass Hg = 2.407x10²⁴ / 6.02x10²³

= 3.998 g of Hg

hope this helps!

5 0

3 years ago

Read 2 more answers

andrew-mc [135]

5 hope this helps I’m pretty sure it should be right

7 0

3 years ago

Example of false solutions and true solutions ​

Example of false solutions and true solutions