Exactly 0.7373 grams of a weak acid (HX) is added to enough water to form 50.00 mL of solution. The resulting solution requires 42.24 mL of 0.1059 M potassium hydroxide (KOH) solution to neutralize the acid. HX + KOH → KX + H₂O Please answer the following questions. A. B. C. D. What is the mass of 1 mole of the weak acid? 1 mol HX = Please provider your answer below. 0² n Determine the equilibrium concentration of HX in the solution after 23.44 mL of 0.1059 M potassium hydroxide (KOH) solution was added to the original weak acid solution. M Please provider your answer below. + 0² $ → → $ (0 grams Determine the equilibrium concentration of KX in the solution after 23.44 mL of 0.1021 M potassium hydroxide (KOH) solution was added to the original weak acid solution. M Please provider your answer below. $ Given that Ka = 3.052×10-10 for the weak acid HX, determine the pH of the solution after 23.44 mL of 0.1059 M potassium hydroxide (KOH) solution was added to the original weak acid solution. pH =

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Exactly 0.7373 grams of a weak acid (HX) is added to enough water to form 50.00 mL of solution. The resulting solution
requires 42.24 mL of 0.1059 M potassium hydroxide (KOH) solution to neutralize the acid.
HX + KOH → KX + H₂O
Please answer the following questions.
A.
B.
C.
D.
What is the mass of 1 mole of the weak acid? 1 mol HX =
Please provider your answer below.
0²
n
Determine the equilibrium concentration of HX in the solution after 23.44 mL of 0.1059 M potassium hydroxide (KOH) solution
was added to the original weak acid solution.
M
Please provider your answer below.
+
0²
$
→
→ $ (0
grams
Determine the equilibrium concentration of KX in the solution after 23.44 mL of 0.1021 M potassium hydroxide (KOH) solution
was added to the original weak acid solution.
M
Please provider your answer below.
$
Given that Ka = 3.052×10-10 for the weak acid HX, determine the pH of the solution after 23.44 mL of 0.1059 M potassium
hydroxide (KOH) solution was added to the original weak acid solution. pH =
Transcribed Image Text:Exactly 0.7373 grams of a weak acid (HX) is added to enough water to form 50.00 mL of solution. The resulting solution requires 42.24 mL of 0.1059 M potassium hydroxide (KOH) solution to neutralize the acid. HX + KOH → KX + H₂O Please answer the following questions. A. B. C. D. What is the mass of 1 mole of the weak acid? 1 mol HX = Please provider your answer below. 0² n Determine the equilibrium concentration of HX in the solution after 23.44 mL of 0.1059 M potassium hydroxide (KOH) solution was added to the original weak acid solution. M Please provider your answer below. + 0² $ → → $ (0 grams Determine the equilibrium concentration of KX in the solution after 23.44 mL of 0.1021 M potassium hydroxide (KOH) solution was added to the original weak acid solution. M Please provider your answer below. $ Given that Ka = 3.052×10-10 for the weak acid HX, determine the pH of the solution after 23.44 mL of 0.1059 M potassium hydroxide (KOH) solution was added to the original weak acid solution. pH =
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