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denis23 [38]
3 years ago
13

For the diprotic weak acid h2a, ka1 = 3.2 × 10-6 and ka2 = 6.1 × 10-9. what is the ph of a 0.0650 m solution of h2a? what are th

e equilibrium concentrations of h2a and a2– in this solution?
Chemistry
1 answer:
Stolb23 [73]3 years ago
5 0
Given:

Diprotic weak acid H2A:
 
Ka1 = 3.2 x 10^-6
Ka2 = 6.1 x 10^-9. 
Concentration = 0.0650 m 

Balanced chemical equation:

H2A ===> 2H+  + A2- 
0.0650       0        0
-x                2x       x
------------------------------
0.065 - x     2x      x

ka1 = 3.2 x 10^-6 = [2x]^2 * [x] / (0.065 - x)

solve for x and determine the concentration at equilibrium. 


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You have 16.7 grams of hydrogen and 15.4 grams of oxygen in a synthesis rxn. Which is the limiting reagent?
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Answer:

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Explanation:

Given data:

Mass of hydrogen = 16.7 g

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Limiting reactant = ?

Solution:

Chemical equation:

2H₂ + O₂   →   2H₂O

Number of moles of hydrogen:

Number of moles = mass/ molar mass

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Number of moles of oxygen:

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Number of moles = 0.48 mol

Now we will compare the moles of both reactant with product,

                 

                             H₂           :          H₂O

                              2            :            2

                             8.35        :            8.35

                             O₂           :          H₂O

                               1            :            2

                             0.48        :        2×0.48 = 0.96 mol

The number of moles of water produced by oxygen are less so it will limiting reactant.

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Answer:

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