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Kisachek [45]
4 years ago
8

Question: In experiment your teacher has provided you with a sample of sandy, salty, ocean water and

Chemistry
1 answer:
liq [111]4 years ago
3 0

Answer:

Explanation:

Homogeneous mixture is a mixture in which the components of the mixture are in the same proportion throughout any sample extracted from the mixture while an heterogeneous mixture is a mixture in which the components of the mixture differ in term of proportion when different samples of the mixture are extracted and compared.

For example, a sandy water will have some parts (usually the bottom) of the mixture with more sand than other parts of the mixture, hence, it (sandy water) is a heterogeneous mixture. While salty and ocean water has it's salt dissolved in the same proportion throughout the water in the mixture, hence salty and/or ocean water is a homogeneous mixture.

Sandy water can be separated by filtration (i.e using a filter paper to separate the sand from the water when the mixture is poured over a filter paper) while salty and ocean water can be separated by distillation (i.e boiling of the mixture so the water molecules can boil and move through a tube as gas or steam into another container where they are cooled and converted back to liquid or water while leaving the solid salt component of the mixture in the boiling tube).

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4 years ago
What is the pH of a buffer solution upon mixing 15.0 mL of 0.40 M HCl and 20.0 mL of 0.50 M NH? Kb (NH3) = 1.8 x 10 E. 7.00 A. 9
vladimir2022 [97]

<u>Answer:</u> The pH of resulting solution is 9.08

<u>Explanation:</u>

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)

  • <u>For HCl:</u>

Molarity of HCl = 0.40 M

Volume of solution = 15.0 mL

Putting values in equation 1, we get:

0.40M=\frac{\text{Moles of HCl}\times 1000}{15.0mL}\\\\\text{Moles of HCl}=0.006mol

  • <u>For ammonia:</u>

Molarity of ammonia = 0.50 M

Volume of solution = 20.0 mL

Putting values in equation 1, we get:

0.50M=\frac{\text{Moles of ammonia}\times 1000}{20.0mL}\\\\\text{Moles of ammonia}=0.01mol

The chemical reaction for hydrochloric acid and ammonia follows the equation:

                  HCl+NH_3\rightarrow NH_4Cl

Initial:          0.006      0.01

Final:             -         0.004              0.006

Volume of solution = 15.0 + 20.0 = 35.0 mL = 0.035 L    (Conversion factor:  1 L = 1000 mL)

  • To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

pOH=pK_b+\log(\frac{[salt]}{[base]})

pOH=pK_b+\log(\frac{[NH_4Cl]}{[NH_3]})

We are given:

pK_b = negative logarithm of base dissociation constant of ammonia = -\log (1.8\times 10^{-5})=4.74

[NH_4Cl]=\frac{0.006}{0.035}

[NH_3]=\frac{0.004}{0.035}

pOH = ?

Putting values in above equation, we get:

pOH=4.74+\log(\frac{0.006/0.035}{0.004/0.035})\\\\pOH=4.92

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

pH+pOH=14\\pH=14-4.92=9.08

Hence, the pH of the solution is 9.08

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