Question

a quantity of $2.00×10^{2}$ ml of 0.862 m hcl is mixed with an equal volume of 0.431 m $ba(oh)_{2}$ in a constant - pressure calorimeter of negligible heat capacity. the initial temperature of the hcl and $ba(oh)_{2}$ solutions is the same at $23.00^{circ}c$. for the process
$h^{+}(aq)+oh^{-}(aq)→h_{2}o(l)$
the heat of neutralization is $-56.2\frac{kj}{mol}$. 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 of \(HCl = 0.862\space M\). Moles of \(H^+ = 0.862\space mol/L\times0.200\space L = 0.1724\space mol\).
Volume of \(Ba(OH)_2 = 0.200\space L\), concentration of \(Ba(OH)_2 = 0.431\space M\). Each \(Ba(OH)_2\) gives 2 \(OH^-\), so moles of \(OH^- = 2\times0.431\space mol/L\times0.200\space L = 0.1724\space mol\).
Both react completely (1:1 ratio in \(H^+ + OH^-
ightarrow H_2O\)). Moles of \(H_2O\) formed = \(0.1724\space mol\).

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

Heat of neutralization is \(-56.2\space kJ/mol\) (negative means exothermic, so heat released \(q = 56.2\space kJ/mol\times0.1724\space mol = 9.68888\space kJ = 9688.88\space J\)).

Step3: Calculate total mass of solution

Density of solution (assume water, \(1\space g/mL\)). Total volume = \(0.200 + 0.200 = 0.400\space L = 400\space mL\). Mass \(m = 400\space g\) (since \(1\space mL\) water ≈ \(1\space g\)). Specific heat \(c = 4.184\space J/g^\circ C\).

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

\(\Delta T = \frac{q}{mc} = \frac{9688.88\space J}{400\space g\times4.184\space J/g^\circ C} \approx 5.74^\circ C\).

Step5: Find final temperature

Initial temperature \(T_i = 23.00^\circ C\). Final temperature \(T_f = T_i + \Delta T = 23.00 + 5.74 = 28.74^\circ C\) (rounded to correct sig figs).

Answer:

\(28.7^\circ C\) (or more precisely \(28.74^\circ C\), but considering sig figs from given data, 3 sig figs in initial temp and concentrations, so \(28.7^\circ C\) or \(28.74^\circ C\) depending on calculation precision; here, precise calculation gives ~\(28.7^\circ C\) to 3 sig figs).

(Note: If we use \(c = 4.18\space J/g^\circ C\) for approximation, \(\Delta T = \frac{9688.88}{400\times4.18} \approx 5.75\), \(T_f = 23 + 5.75 = 28.75^\circ C\), but with exact steps, ~\(28.7^\circ C\) to 3 sig figs.)

Final answer (with correct sig figs): \(\boxed{28.7^\circ C}\) (or \(28.74^\circ C\) if more precision, but typically 3 sig figs from data gives \(28.7^\circ C\)).