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Anna11 [10]
4 years ago
12

Phosphoric acid is a triprotic acid ( K a1 = 6.9 × 10 − 3 , K a2 = 6.2 × 10 − 8 , and K a3 = 4.8 × 10 − 13 ). To find the pH of

a buffer composed of H 2 PO − 4 ( aq ) and HPO 2 − 4 ( aq ) , which p K a value should be used in the Henderson–Hasselbalch equation?
Chemistry
1 answer:
timama [110]4 years ago
7 0

Answer:

Ka2 values should be used in obtaining pKa2 in the Henderson–Hasselbalch equation

Explanation:

During first dissociation from H3PO4 to H2PO4-, ka1 will be used, and from H2PO4-to HPO4-, ka2 values will be used, hence a buffer solution containing H2PO4-and HPO4-will use ka2 in obtaining pKa2 in the Henderson–Hasselbalch equation.

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What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?​
UNO [17]

Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                      NH₃          :            N₂

                        4             :             2

                      0.12           :           2/4×0.12 = 0.06

                      NH₃         :            H₂O

                        4            :             6

                        0.12       :           6/4×0.12 = 0.18

                       

                       O₂            :            N₂

                        3             :             2

                      0.125        :           2/3×0.125 = 0.08

                        O₂           :            H₂O

                        3              :             6

                        0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                        NH₃          :             O₂

                           4            :              3

                           0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

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[tex]\purple{\rule{45pt}{7pt}}\purple{\rule{45pt}{999999pt}}[tex]

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3 years ago
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