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Ghella [55]
3 years ago
5

Calculate the ph of a solution containing 0.0451 m potassium hydrogen tartrate and 0.028 m dipotassium tartrate. The ka values f

or tartaric acid are 9.20 × 10-4 (ka1) and 4.31 × 10-5 (ka2).
Chemistry
1 answer:
Marina86 [1]3 years ago
4 0

Given buffer:

potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )

[KHC4H4O6] = 0.0451 M

[K2C4H4O6] = 0.028 M

Ka1 = 9.2 *10^-4

Ka2 = 4.31*10^-5

Based on Henderson-Hasselbalch equation;

pH = pKa + log [conjugate base]/[acid]

where pka = -logKa

In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2

pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]

     = -log (4.31*10^-5) + log [0.0451]/[0.028]

pH = 4.15



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Answer:

<h2>mass = 29.25 g</h2>

Explanation:

The denisty of a substance can be found by using the formula

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What is the difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm
Zanzabum

Answer:

The difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm is :

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This means, 1 mole of gold(Au) contain = 196.96 grams

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<u>Part II :</u>

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Volume of the cuboid = length\times breadth\times height

Volume of the gold bar =6.00\times 4.25\times 2.00

Volume of the gold bar = 51cm^{3}

Using formula,

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Mass = Density\times Volume

Mass = 19.32 \times 51

Mass = 985.32 g

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<u>Difference</u>  <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

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