0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.
Potassium hydrogen phthalate, KHP, is a monoprotic acid often used to standardize NaOH solutions.
The balanced neutralization equation is:
NaOH(aq) + KHC₈H₄O₄(aq) ⇒ KNaC₈H₄O₄(aq) + H₂O(l)
- Step 1: Calculate the reacting moles of KHP.
0.212 g of KHP react. The molar mass of KHP is 204.22 g/mol.
0.212 g × 1 mol/204.22 g = 1.04 × 10⁻³ mol
- Step 2: Determine the reacting moles of NaOH.
The molar ratio of NaOH to KHP is 1:1.
1.04 × 10⁻³ mol KHP × 1 mol NaOH/1 mol KHP = 1.04 × 10⁻³ mol NaOH
- Step 3: Calculate the molarity of NaOH.
1.04 × 10⁻³ moles of NaOH are in 35.00 mL of solution.
[NaOH] = 1.04 × 10⁻³ mol / 35.00 × 10⁻³ L = 0.0297 M
0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.
Learn more about titration here: brainly.com/question/4225093
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Answer:
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
Explanation:
Step 1: Data given
The combustion reaction of octane produces 5104.1 kJ per mol octane
Step 2: The balanced equation
C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g) ∆H°rxn = -5104.1 kJ/mol
Step 3:
∆H°rxn = ∆H°f of products minus the ∆H° of reactants
∆H°rxn = ∆H°f products - [∆H°f reactants]
-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)
∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol
∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ
/mol
∆H°f C8H18 = -220.1 kJ/mol
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol