Question

a quantity of (2.00\times10^{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 (22.61^{circ}c). for the process
(h^{+}(aq)+oh^{-}(aq)\to 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 = \(0.862\space M\), so 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, moles of \(H_2O\) formed = \(0.1724\space mol\).

Step2: Calculate heat released (q)

Heat of neutralization = \(-56.2\space kJ/mol\), so heat released \(q = 56.2\space kJ/mol\times0.1724\space mol = 9.69\space kJ = 9690\space J\) (since reaction is exothermic, q is positive for solution).

Step3: Calculate total mass of solution

Density of solution (assume same as 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 \(c = 4.184\space J/g^\circ C\).
\(\Delta T=\frac{q}{mc}=\frac{9690\space J}{400\space g\times4.184\space J/g^\circ C}\approx5.73^\circ C\).

Step5: Calculate final temperature

Initial temperature \(T_i = 22.61^\circ C\), final temperature \(T_f = T_i+\Delta T = 22.61 + 5.73 = 28.34^\circ C\).

Answer:

\(28.3^\circ C\) (or more precisely \(28.34^\circ C\) depending on sig figs, but 3 sig figs from given data: 2.00×10², 0.862, 0.431, 22.61, 56.2 – so 3 or 4? Wait, moles calculation: 0.200 L (3 sig figs), 0.862 (3), 0.431 (3). So final temp: 28.3 °C (or 28.34, but let's check calculation again. Wait, q = 56.20.1724 = 56.20.1724: 56.20.1724 = 56.20.1 + 56.20.07 + 56.20.0024 = 5.62 + 3.934 + 0.1349 = 9.6889 kJ = 9688.9 J. Then \(\Delta T = 9688.9/(4004.184) = 9688.9/1673.6 ≈ 5.79\)? Wait, 4004.184=1673.6. 9688.9/1673.6≈5.79. Then Tf=22.61+5.79=28.40? Wait, maybe I messed up moles. Wait, \(Ba(OH)_2\) volume: 2.00×10² mL is 200 mL, so equal volume is 200 mL. So total volume 400 mL, mass 400 g (density 1 g/mL). Moles of \(OH^-\): 0.431 M 0.2 L 2 = 0.1724 mol. Moles of \(H^+\): 0.862 M 0.2 L = 0.1724 mol. So moles of reaction: 0.1724 mol. Heat: 56.2 kJ/mol 0.1724 mol = 56.20.1724. Let's calculate 0.172456.2: 0.156.2=5.62, 0.0756.2=3.934, 0.002456.2=0.13488. Sum: 5.62+3.934=9.554+0.13488=9.68888 kJ=9688.88 J. Then \(\Delta T = q/(mc)\). m=400 g, c=4.184 J/g°C. So 9688.88/(4004.184)=9688.88/1673.6≈5.79°C. Then Tf=22.61+5.79=28.40°C. Wait, maybe my initial \(\Delta T\) was wrong. Let's recalculate: 4004.184=1673.6. 9688.88/1673.6=5.79. So 22.61+5.79=28.40, which rounds to 28.4 °C? Wait, but let's check significant figures. The given data: 2.00×10² (3 sig figs), 0.862 (3), 0.431 (3), 22.61 (4), 56.2 (3). So the least number of sig figs in multiplication/division is 3. So moles: 0.200 L (3), 0.862 (3), 0.431 (3) – so moles are 0.172 (3 sig figs? Wait, 0.8620.200=0.1724, which is 0.172 (3 sig figs? No, 0.200 has 3, 0.862 has 3, so 0.1724 is 0.172 (3 sig figs? Wait, 0.200 is 3, 0.862 is 3, so 0.8620.200=0.1724, which should be 0.172 (3 sig figs). Similarly, 0.4310.2002=0.1724, 0.172. Then heat: 56.20.172=56.20.172. 56.20.1=5.62, 56.20.07=3.934, 56.20.002=1.124? Wait no, 0.172=0.1+0.07+0.002? No, 0.172=0.17+0.002? Wait, 0.172 is 0.1 + 0.07 + 0.002? No, 0.172=0.17 + 0.002? Anyway, 56.20.172=9.6664 kJ=9666.4 J. Then \(\Delta T=9666.4/(4004.184)=9666.4/1673.6=5.776\). Then 22.61+5.776=28.386, which is 28.4 °C (3 sig figs) or 28.39 (4). But maybe the correct answer is 28.3 or 28.4. Wait, let's check the calculation again.

Wait, the key steps:

1. Moles of H⁺: 0.862 M * 0.200 L = 0.1724 mol (correct, 0.200 has 3 sig figs, 0.862 has 3, so 0.172 mol? No, 0.200 is 3, 0.862 is 3, so 0.1724 is 0.172 (3 sig figs) when rounded? Wait, no, 0.200 is 3, 0.862 is 3, so the product is 0.1724, which should be 0.172 (3 sig figs) for intermediate steps? Or keep more digits for accuracy.

1. Heat: 56.2 kJ/mol * 0.1724 mol = 9.68888 kJ = 9688.88 J.

1. Mass: 400 g (since 400 mL, density 1 g/mL).

1. \(q = mc\Delta T\) → \(\Delta T = q/(mc) = 9688.88 / (400 * 4.184) = 9688.88 / 1673.6 ≈ 5.79\) °C.

1. Final temp: 22.61 + 5.79 = 28.40 °C. Rounding to 3 sig figs (from 56.2, 0.862, 0.431, 2.00×10²), so 28.4 °C. But maybe the answer is 28.3 °C? Wait, let's check with exact calculation:

\(c = 4.184\space J/g^\circ C\) (specific heat of water).

\(q = 56.2\space kJ/mol \times n\), where \(n\) is moles of reaction.

Moles of \(H^+\): \(0.862\space M \times 0.200\space L = 0.1724\space mol\).

Moles of \(OH^-\): \(2 \times 0.431\space M \times 0.200\space L = 0.1724\space mol\).

So \(n = 0.1724\space mol\).

\(q = 56.2\space kJ/mol \times 0.1724\space mol = 9.68888\space kJ = 9688.88\space J\).

\(m = (200\space mL + 200\space mL) \times 1.00\space g/mL = 400\space g\).

\(\Delta T = \frac{q}{mc} = \frac{9688.88\space J}{400\space g \times 4.184\space J/g^\circ C} = \frac{9688.88}{1673.6} ≈ 5.79^\circ C\).

\(T_f = 22.61^\circ C + 5.79^\circ C = 28.40^\circ C\).

Considering significant figures, 56.2 has 3, 0.862 has 3, 0.431 has 3, 2.00×10² has 3, 22.61 has 4. The least number of significant figures in the multiplication/division steps is 3, so the final answer should have 3 significant figures. 28.40 rounds to 28.4 °C.