Question

concentration of hydrobromic acid (m) 2.01
concentration of potassium hydroxide (m) 2.15
calorimeter constant (j/°c) 14.9
volume of hydrobromic acid (ml) 93.0
temperature of hydrobromic acid (°c) 25.0
volume of potassium hydroxide (ml) 96.5
temperature of potassium hydroxide (°c) 25.0
final temperature after mixing (°c) 38.7
mass of mixed solution (g) 189.5
moles of h⁺ (mol)
moles of oh⁻ (mol)
limiting reactant
δhneutralization (kj/mol)
how to calculate the enthalpy of neutralization

Explanation:

Step1: Calculate moles of \(H^+\)

Use formula \(n = C\times V\) (where \(n\) is moles, \(C\) is concentration, \(V\) is volume in liters).
\(V_{HBr}=93.0\space mL = 0.093\space L\), \(C_{HBr}=2.01\space M\)
\(n_{H^+}=2.01\times0.093 = 0.18693\space mol\)

Step2: Calculate moles of \(OH^-\)

\(V_{KOH}=96.5\space mL=0.0965\space L\), \(C_{KOH} = 2.15\space M\)
\(n_{OH^-}=2.15\times0.0965=0.207475\space mol\)

Step3: Determine limiting reactant

Since \(n_{H^+}

Step4: Calculate heat released \(q\)

Use \(q=(m\times c\times\Delta T)+(C_{cal}\times\Delta T)\), assume \(c = 4.18\space J/(g\cdot^{\circ}C)\), \(\Delta T=38.7 - 25.0=13.7^{\circ}C\), \(m = 189.5\space g\), \(C_{cal}=14.9\space J/^{\circ}C\)
\(q=(189.5\times4.18\times13.7)+(14.9\times13.7)\)
\(q=(189.5\times4.18\times13.7)+204.13\)
\(q=(801.11\times13.7)+204.13\)
\(q = 10975.207+204.13=11179.337\space J\approx11.18\space kJ\)

Step5: Calculate \(\Delta H_{neutralization}\)

\(\Delta H=\frac{-q}{n_{limiting}}\) (negative because exothermic)
\(\Delta H=\frac{- 11.18}{0.18693}\approx - 59.8\space kJ/mol\)

Answer:

Moles of \(H^+\): \(0.187\space mol\); Moles of \(OH^-\): \(0.207\space mol\); Limiting Reactant: \(H^+\); \(\Delta H_{neutralization}\approx - 59.8\space kJ/mol\)