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 $21.38 \\, ^\circ\text{c}$. 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 reactants

Volume of HCl = \(2.00\times10^{2}\space mL = 0.200\space L\), molarity of HCl = \(0.862\space M\). Moles of \(H^+\) (from HCl) = \(M\times V = 0.862\space mol/L\times0.200\space L = 0.1724\space mol\).
Volume of \(Ba(OH)_2\) = \(0.200\space L\), molarity of \(Ba(OH)_2 = 0.431\space M\). Moles of \(OH^-\) (from \(Ba(OH)_2\)) = \(2\times0.431\space mol/L\times0.200\space L = 0.1724\space mol\). So, moles of \(H_2O\) formed = \(0.1724\space mol\) (since \(H^+\) and \(OH^-\) react 1:1).

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

Heat of neutralization (\(\Delta H\)) = \(-56.2\space kJ/mol\), so heat released (\(q = -\Delta H\times n\)) = \(56.2\space kJ/mol\times0.1724\space mol = 9.68888\space kJ = 9688.88\space J\).

Step3: Calculate total mass of solution

Density of solution (assume same as 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 (\(c\)) of water = \(4.184\space J/g^\circ C\).

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

\(\Delta T=\frac{q}{mc}=\frac{9688.88\space J}{400\space g\times4.184\space J/g^\circ C}\approx5.72^\circ C\).

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

Initial temperature (\(T_i\)) = \(21.38^\circ C\). \(T_f = T_i + \Delta T = 21.38^\circ C + 5.72^\circ C = 27.10^\circ C\) (after considering significant digits, check calculations for precision). Wait, recalculate \(\Delta T\): \(q = 9688.88\space J\), \(m = 400\space g\), \(c = 4.184\space J/g^\circ C\). \(\Delta T=\frac{9688.88}{400\times4.184}=\frac{9688.88}{1673.6}\approx5.79^\circ C\). Then \(T_f = 21.38 + 5.79 = 27.17^\circ C\)? Wait, let's recheck moles:
Wait, moles of \(H^+\): \(0.862\times0.2 = 0.1724\), moles of \(OH^-\): \(2\times0.431\times0.2 = 0.1724\), so moles of reaction = 0.1724. Heat released: \(0.1724\times56200\space J = 0.1724\times56200 = 9688.88\space J\). Mass: 400g, \(c = 4.184\). \(\Delta T = 9688.88/(400\times4.184) = 9688.88/1673.6 ≈ 5.79^\circ C\). So \(T_f = 21.38 + 5.79 = 27.17^\circ C\)? Wait, maybe I made a mistake in \(\Delta T\) calculation. Wait, \(q\) is positive (heat absorbed by solution), so \(\Delta T = T_f - T_i\), so \(T_f = T_i + \Delta T\). Let's recalculate \(\Delta T\): \(9688.88\space J / (400\space g\times4.184\space J/g^\circ C) = 9688.88 / 1673.6 ≈ 5.79^\circ C\). So \(T_f = 21.38 + 5.79 = 27.17^\circ C\)? Wait, but let's check significant digits. The given values: 2.00×10² (3 sig figs), 0.862 (3), 0.431 (3), 21.38 (4), 56.2 (3). So the answer should have 3 or 4? Wait, moles: 0.1724 (from 0.862×0.200: 0.862 has 3, 0.200 has 3, so 0.172 (3 sig figs? Wait, 0.862×0.200 = 0.1724, but 0.200 is 3 sig figs, 0.862 is 3, so 0.172 (3 sig figs). Wait, maybe I messed up. Let's redo moles:

Moles of HCl: \(0.862\space M\times0.200\space L = 0.1724\space mol\) (0.200 is 3 sig figs, 0.862 is 3, so 0.172 mol? Wait, 0.200 is 3, 0.862 is 3, so product is 3 sig figs: 0.172 mol. Similarly, \(Ba(OH)_2\): 0.431×0.200×2 = 0.1724, same. So moles of reaction: 0.172 mol. Then heat: 56.2×0.172 = 9.6664 kJ = 9666.4 J. Then \(\Delta T = 9666.4 / (400×4.184) = 9666.4 / 1673.6 ≈ 5.777^\circ C\). Then \(T_f = 21.38 + 5.777 ≈ 27.16^\circ C\), which rounds to 27.2^\circ C? Wait, no, maybe my initial moles were wrong. Wait, HCl is 0.862 M, 0.200 L: 0.862×0.200 = 0.1724 mol (correct, 0.200 has 3 decimal places? No, 0.200 L is 200 mL, which is 3 sig figs). \(Ba(OH)_2\) is 0.431 M, 0.200 L: moles of \(Ba(OH)_2 = 0.431×0.200 = 0.0862\) mol, so moles of \(OH^- = 2×0.0862 = 0.1724\) mol. So H+ and OH- are equal, so moles of water: 0.1724 mol. So heat: 56.2 kJ/mol × 0.1724 mol = 9.68888 kJ = 9688.88 J. Mass: 400 g (200 + 200 mL, density 1 g/mL, so 400 g, 3 sig figs? 200 mL is 3 sig figs, so 400 g is 3 sig figs). Specific heat: 4.184 J/g°C (4 sig figs). So \(\Delta T = 9688.88 / (400×4.184) = 9688.88 / 1673.6 ≈ 5.79°C\). Then Tf = 21.38 + 5.79 = 27.17°C, which with significant digits: the least number of sig figs in the multiplication/division steps: 56.2 (3), 0.1724 (4, but from 0.862 (3) and 0.200 (3), so 0.172 (3)), 400 (3), 4.184 (4). So the \(\Delta T\) should have 3 sig figs: 5.80°C? Wait, no, 9688.88 / 1673.6 ≈ 5.79, which is ~5.80 with 3 sig figs. Then Tf = 21.38 + 5.79 = 27.17, which is 27.2°C? Wait, maybe I made a mistake in mass. Wait, total volume is 400 mL, so mass is 400 g (since density of solution is assumed 1 g/mL, which is an approximation). So that's correct.

Wait, let's check again:

1. Moles of H+: 0.862 M * 0.200 L = 0.1724 mol (0.200 is 3 sig figs, 0.862 is 3, so 0.172 mol? Wait, 0.862 * 0.200 = 0.1724, but 0.200 has 3, so 0.172 (3 sig figs).
2. Moles of OH-: 2 * 0.431 M * 0.200 L = 0.1724 mol (same as above, 0.431 is 3, 0.200 is 3, so 0.172 mol).
3. Heat released: 56.2 kJ/mol * 0.1724 mol = 9.68888 kJ = 9688.88 J (56.2 has 3, 0.1724 has 4, so 9.69 kJ (3 sig figs) = 9690 J).
4. Mass: 400 g (3 sig figs).
5. Specific heat: 4.184 J/g°C (4 sig figs).
6. \(\Delta T = q / (m*c) = 9690 J / (400 g * 4.184 J/g°C) = 9690 / 1673.6 ≈ 5.79°C\) (5.80 with 3 sig figs).
7. Final temperature: 21.38°C + 5.79°C = 27.17°C, which rounds to 27.2°C (3 sig figs) or 27.17°C (4 sig figs). But the initial temperature is 21.38 (4 sig figs), heat of neutralization 56.2 (3), volumes 2.00×10² (3), 0.862 (3), 0.431 (3). So the answer should have 3 sig figs: 27.2°C? Wait, no, 21.38 is 4, but the least number of sig figs in the multiplication steps is 3 (from 56.2, 0.862, 0.431, 2.00×10²). So the final answer should have 3 sig figs: 27.2°C? Wait, but when I calculated \(\Delta T\) as 5.79, adding to 21.38 gives 27.17, which is 27.2 when rounded to 3 sig figs.

Answer:

\(\boxed{27.2}\) (or more precisely, 27.17, but with significant digits, 27.2 °C)