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DENIUS [597]
2 years ago
11

Isaac pour the remaining chemical into the sink and apparatus were left in the sink​

Chemistry
2 answers:
adelina 88 [10]2 years ago
8 0
This is not a good idea u never know what the chemicals could do when in contact with metal or pipes or even water. always ask your chem teacher and make sure they tell u before. also alert them if this has happened because they can alert the school and act as needed
skelet666 [1.2K]2 years ago
3 0

Answer:

That's a major lab mistake! Never pour chemicals in sink unless otherwise stated by the lab instructor. They should go in a properly labelled waste jar or similar container. Apparatus should not be left in sink, it should be washed and cleaned, and returned to where it came from.

Explanation:

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If you mix 50mL of 0.1 M TRIS acid with 60 mL of0.2 M<br> TRIS base, what will be the resulting pH?
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<u>Answer:</u> The pH of resulting solution is 8.7

<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)}}

  • <u>For TRIS acid:</u>

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol

  • <u>For TRIS base:</u>

Molarity of TRIS base solution = 0.2 M

Volume of solution = 60 mL

Putting values in above equation, we get:

0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol

Volume of solution = 50 + 60 = 110 mL = 0.11 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{[salt]}{[acid]})

pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})

We are given:

pK_a = negative logarithm of acid dissociation constant of TRIS acid = 8.3

[\text{TRIS acid}]=\frac{0.005}{0.11}

[\text{TRIS base}]=\frac{0.012}{0.11}

pH = ?

Putting values in above equation, we get:

pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7

Hence, the pH of resulting solution is 8.7

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3 years ago
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