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quester [9]
3 years ago
6

A 500.0 ml buffer solution is 0.10 m benzoic acid and 0.10 m sodium benzoate has an initial ph of 4.19. what is the ph of the bu

ffersolution upon addition of 0.010 mol ofnaoh? the kafor benzoic acid is 6.5 • 10-5
Chemistry
1 answer:
Delvig [45]3 years ago
5 0
New pH = 4.27

Adding 0.01 mol of NaOH would increase the amount (in mol, 0.10) of sodium benzoate and decrease the amount (in mol, 0.10) of benzoic acid. This is because NaOH would react with benzoic acid to form more sodium benzoate. NaOH would get used up in the solution. Adding (sodium benzoate + NaOH) and subtracting (benzoic acid + NaOH) would give you 0.11 mol of sodium benzoate and 0.09 mol of benzoic acid. New pH = pKa + log [mol of sodium benzoate/mol of benzoic acid]. Hope this helps! 
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Answer:

it would need to lose energy

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3 years ago
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USPshnik [31]

Answer:

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3 years ago
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pshichka [43]

Answer:

Explanation:

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= 35.96755 x .00337 + 37.96272 x .00063 + 39.96240 x .99600

= .12121 + .0239165 + 39.80255

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27.97693 x .9223 + 28.97649 x .0467 + 29.97376 x .0310

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b )

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= 16.76 x 10²³ .

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2 )

C₁₆H₁₅F₂N₃O₄S

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= 383

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7 0
3 years ago
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kompoz [17]

Answer:

–1

Explanation:

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2 8 7

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Therefore the charge on the ion of the element will be –1 indicating that the atom has received 1 electron to complete it's octet configuration.

5 0
3 years ago
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GrogVix [38]
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