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anygoal [31]
2 years ago
15

Determine whether the stopcock should be completely open, partially open, or completely closed for each activity involved with t

itration.
Close to the calculated endpoint of a titration ________
At the beginning of a titration _______
Filling the buret with titrant ________
Conditioning the buret with titrant _______
Chemistry
1 answer:
densk [106]2 years ago
3 0

Answer:

Close to the calculated endpoint of a titration - <u>Partially open</u>

At the beginning of a titration - <u>Completely open</u>

Filling the buret with titrant - <u>Completely closed</u>

Conditioning the buret with the titrant - <u>Completely closed</u>

Explanation:

'Titration' is depicted as the process under which the concentration of some substances in a solution is determined by adding measured amounts of some other substance until a rection is displayed to be complete.

As per the question, the stopcock would remain completely open when the process of titration starts. After the buret is successfully placed, the titrant is carefully put through the buret in the stopcock which is entirely closed. Thereafter, when the titrant and the buret are conditioned, the stopcock must remain closed for correct results. Then, when the process is near the estimated end-point and the solution begins to turn its color, the stopcock would be slightly open before the reading of the endpoint for adding the drops of titrant for final observation.

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A 500 mL sample of a 0.100 M formate buffer, pH 3.75, is treated with 5 mL of 1.00 M KOH. What is the pH following this addition
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<u>Answer:</u> The pH of the solution after addition of KOH is 3.84

<u>Explanation:</u>

We are given:

pH of buffer = 3.75

pK_a of formic acid = 3.75

Using Henderson-Hasselbalch equation for formate buffer:

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

Putting values in above equation, we get:

3.75=3.75+\log(\frac{[HCOO-]}{[HCOOH]})\\\\\log(\frac{[HCOO-]}{[HCOOH]})=0\\\\\frac{[HCOO-]}{[HCOOH]}=1

[HCOO-]=[HCOOH]

We are given:

Concentration of formate buffer = 0.100 M

[HCOO-]+[HCOOH]=0.1

[HCOO-]=[HCOOH]=0.05M

As, the volume of buffer is the same. So, the concentration is taken as number of moles of formate ions as well as formic acid

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)}}

Molarity of KOH = 1.00 M

Volume of solution = 5 mL

Putting values in above equation, we get:

1.00M=\frac{\text{Moles of KOH}\times 1000}{5mL}\\\\\text{Moles of KOH}=0.005mol

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

                  HCOOH+KOH\rightarrow HCOO^-+H_2O

Initial:       0.05    0.005               0.05

Final:         0.045          -                0.055          

Volume of solution = 500 + 5 = 505 mL = 0.505 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{[HCOO^-]}{[HCOOH]})

We are given:

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

[HCOO^-]=\frac{0.055}{0.505}

[HCOOH]=\frac{0.045}{0.505}

pH = ?

Putting values in above equation, we get:

pH=3.75+\log(\frac{0.055/0.505}{0.045/0.505})\\\\pH=3.84

Hence, the pH of the solution after addition of KOH is 3.84

3 0
2 years ago
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Since the Carbon C is 17.39% by mass hence the Fluorine F is 82.61% by mass. Divide each mass % by the respective molar masses, that is:

 

C = 17.39 / 12 = 1.45

F = 82.61 / 19 = 4.35

 

Divide the two by the smaller number, so divide by 1.45

 

C = 1.45 / 1.45 = 1

F = 4.35 / 1.45 = 3

 

So the empirical formula is:

CF3 

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