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Ket [755]
3 years ago
8

a 0.5678 of KHP required 26.64cm³ of NaOH to complete neutralization.calculate the molarity of the NaOH solution​

Chemistry
1 answer:
vladimir2022 [97]3 years ago
4 0

Answer:

Explanation:

0.5678 G        X GRAMS

KHC8H4O4 + NaOH = NaKC8H4O4 + H2O

1 MOL               1 MOL

0.5678G X 204G/MOL = 0.00278 MOL KHC8H4O4

0.00278 MOL KHC8H4O4 X 1 MOLE NaOH/1 MOLE  KHC8H4O4=0.00278 MOL NaOH

0.00278 MOL NaOH/26.26ml=0.106 molar

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During the titration of a diluted vinegar sample with a sodium hydroxide solution, the volume of sodium hydroxide used was less
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Answer:

B and D could be true

Explanation:

A volume of sodium hydroxide less than expected could occurs for two reasons:

The real concentration of sodium hydroxide was higher than expected or the amount of vinegar added was less than expected:

A. The sodium hydroxide solution had been allowed to stand exposed to the air for a long time prior to the titration.  FALSE. A long expose  to the air decreases concentration of the NaOH.

B. The volumetric flask used to prepare the diluted vinegar solution was rinsed with water prior to use.  TRUE. You add a less amount of vinegar doing you require less amount of NaOH than expected.

C. The burette used to deliver the sodium hydroxide solution was rinsed with water prior to use.  FALSE. Thus, you add a less amount of NaOH than expected. To explain the matter, you add more NaOH than expected.

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denis23 [38]

Answer:

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5 0
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ruslelena [56]

Answer:

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Explanation:

3 0
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What is the pH of a mixture of 0.042 M NaH2PO4 and 0.058 M Na2HPO4? Hint: The pKa of phosphate is 6.86.
AlekseyPX

Answer:

The pH value of the mixture will be 7.00

Explanation:

Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,

pH=pK_{a} + log(\frac{[Base]}{[Acid]})

According to the given conditions, the equation will become as follow

pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})

The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.

Placing all the given data we obtain,

pH=6.86 + log(\frac{0.058}{0.042})

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Vlad1618 [11]

Answer:

False

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