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KengaRu [80]
3 years ago
10

The ka of phosphoric acid, h3po4, is 7.6  10–3 at 25 °c. for the reaction h3po4(aq) h2po4 – (aq) + h+ (aq) ∆h° = –14.2 kj/mol.

what is the ka of h3po4 at 60 °c?
Chemistry
1 answer:
blsea [12.9K]3 years ago
7 0
<span>Use the van't Hoff equation: ln ( K2 K1 ) = Δ Hº R ( 1 T1 ⒠1 T2 ) ln ( K2 7.6*10^-3 ) = -14,200 J 8.314 ( 1 298 ⒠1 333 ) ln ( K2 7.6*10^-3 ) = ⒠1708 ( 0.00035 ) ln ( K2 0.0076 ) = ⒠0.598 Apply log rule a = log b b a -0.598 = ln ( e ⒠0.598 ) = ln ( 1 e 0.598 ) Multiply both sides with e^0.598 K 2 e 0.598 = 0.0076 K e 0.598 e 0.598 = 0.0076 e 0.598 K 2 = 0.0076 e 0.598 = 4.2 ⋅ 10 ⒠3 K2 = 4.2 ⋅ 10 ⒠3</span>
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<h3>Answer:</h3>

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<h3>Explanation:</h3>

We have;

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We are required to determine the mass of 9.51 g of a NaBr sample.

  • Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.

In this case,

  • A sample of 11.97 g of NaBr contains 22.34% of Na by mass
  • Therefore;

A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass

  • But;

% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100

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Mass of the element = (% composition of an element × mass of the compound) ÷ 100

Therefore;

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Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g

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