Question

a quantity of $2.00 \times 10^2$ ml of 0.862 m hcl is mixed with an equal volume of 0.431 m $\ce{ba(oh)2}$ in a constant-pressure calorimeter of negligible heat capacity. the initial temperature of the hcl and $\ce{ba(oh)2}$ solutions is the same at 20.96 $^\circ$c. for the process\
\
$\ce{h^+(aq) + oh^-(aq) -> h2o(l)}$\
\
the heat of neutralization is $-56.2 \frac{\text{kj}}{\text{mol}}$. what is the final temperature of the mixed solution? be sure your answer has the correct number of significant digits.\
note: reference the phase change properties of pure substances table for additional information.

Explanation:

Step1: Calculate moles of \( H^+ \) and \( OH^- \)

Volume of \( HCl = 2.00\times10^2\space mL = 0.200\space L \), concentration \( [HCl] = 0.862\space M \). Moles of \( H^+ = 0.200\space L\times0.862\space mol/L = 0.1724\space mol \).
Volume of \( Ba(OH)_2 = 0.200\space L \), concentration \( [Ba(OH)_2] = 0.431\space M \). Moles of \( OH^- = 2\times0.200\space L\times0.431\space mol/L = 0.1724\space mol \).

Step2: Calculate heat released (\( q \))

Reaction: \( H^+(aq) + OH^-(aq)
ightarrow H_2O(l) \), \( \Delta H = -56.2\space kJ/mol \).
Moles of reaction = moles of \( H^+ \) (or \( OH^- \)) = \( 0.1724\space mol \).
\( q = \Delta H\times n = -56.2\space kJ/mol\times(-0.1724\space mol) = 9.69\space kJ = 9690\space J \) (heat absorbed by solution, so \( q = +9690\space J \)).

Step3: Calculate total mass of solution

Density of solution ≈ density of water = \( 1.00\space g/mL \). Total volume = \( 0.200 + 0.200 = 0.400\space L = 400\space mL \).
Mass \( m = 400\space mL\times1.00\space g/mL = 400\space g \).

Step4: Use \( q = mc\Delta T \) to find \( \Delta T \)

Specific heat of water \( c = 4.184\space J/g^\circ C \).
\( \Delta T = \frac{q}{mc} = \frac{9690\space J}{400\space g\times4.184\space J/g^\circ C} \approx 5.74^\circ C \).

Step5: Calculate final temperature (\( T_f \))

Initial temperature \( T_i = 20.96^\circ C \).
\( T_f = T_i + \Delta T = 20.96 + 5.74 = 26.70^\circ C \).

Answer:

\( 26.7^\circ C \) (or \( 26.70^\circ C \) with proper sig figs)