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 ( 21.62^{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 = \(2.00\times10^{2}\,mL = 0.200\,L\), concentration of HCl = \(0.862\,M\). Moles of \(H^+\) = \(M\times V = 0.862\,mol/L\times0.200\,L = 0.1724\,mol\).
Volume of \(Ba(OH)_2\) = \(0.200\,L\), concentration of \(Ba(OH)_2 = 0.431\,M\). Each \(Ba(OH)_2\) gives 2 \(OH^-\), so moles of \(OH^-\) = \(2\times0.431\,mol/L\times0.200\,L = 0.1724\,mol\).
Both react completely (1:1 stoichiometry in \(H^+ + OH^- ightarrow H_2O\)). Moles of \(H_2O\) formed = \(0.1724\,mol\).

Step2: Calculate heat released (q)

Heat of neutralization = \(-56.2\,\frac{kJ}{mol}\), so heat released \(q = \vert\Delta H\vert\times n = 56.2\,kJ/mol\times0.1724\,mol = 9.68888\,kJ = 9688.88\,J\) (since \(1\,kJ = 1000\,J\)).

Step3: Calculate total mass of solution

Density of solution (assume same as water) = \(1.00\,g/mL\). Total volume = \(200\,mL + 200\,mL = 400\,mL\). Mass \(m = ho\times V = 1.00\,g/mL\times400\,mL = 400\,g\). Specific heat of water \(c = 4.184\,J/(g\cdot^\circ C)\).

Step4: Relate heat to temperature change (\(q = mc\Delta T\))

\(q = mc(T_f - T_i)\), so \(T_f = T_i + \frac{q}{mc}\).
Substitute values: \(T_f = 21.62^\circ C + \frac{9688.88\,J}{400\,g\times4.184\,J/(g\cdot^\circ C)}\).
Calculate \(\frac{9688.88}{400\times4.184} = \frac{9688.88}{1673.6} \approx 5.79^\circ C\).
\(T_f = 21.62 + 5.79 = 27.41^\circ C\) (check significant digits: moles from 3 sig figs in volume, 3 in conc; heat has 3 sig figs. Final temp should have 3 decimal? Wait, initial temp 21.62 (4 sig figs), but calculations: moles 0.1724 (4), heat 9.688 (4), mass 400 (3? Wait, volume 2.00×10² (3 sig figs), so total volume 400 (3 sig figs), mass 400 (3). But let's recheck:

Wait, volume of HCl: \(2.00\times10^2\,mL\) (3 sig figs), so 0.200 L (3). Concentration 0.862 (3), so moles \(H^+\): \(0.862\times0.200 = 0.1724\) (but 3 sig figs? Wait, 2.00×10² is 3, 0.862 is 3, so 0.172 (3 sig figs? Wait no: 2.00 has 3, 0.862 has 3, so 0.862×0.200 = 0.1724 (but 3 sig figs? Wait, 0.200 is 3, so 0.862×0.200 = 0.1724, but we take 0.172 (3)? Wait no, 2.00×10² is 3, 0.862 is 3, so 0.862*0.200 = 0.1724 (the 0.200 has 3, so the product has 3? Wait, 0.200 is 3 sig figs, 0.862 is 3, so 0.862×0.200 = 0.1724, but we should keep 3 sig figs? Wait, no: 2.00×10² mL is 3 sig figs, so 0.200 L (3). 0.862 M (3). So moles H⁺: 0.862 * 0.200 = 0.1724 (but 3 sig figs? Wait, 0.200 is 3, so 0.862*0.200 = 0.1724, but the limiting is 3, so 0.172 mol? Wait, no, 2.00×10² is 3, 0.862 is 3, so 0.862*0.200 = 0.1724 (the 0.200 has 3, so the result should have 3? Wait, 0.200 is 3, 0.862 is 3, so 3×3=3 sig figs. So 0.172 mol. But earlier calculation with 0.1724 is okay for intermediate steps.

But let's proceed with the numbers:

q = 56.2 kJ/mol * 0.1724 mol = 9.688 kJ = 9688 J.

m = 400 g (since 400 mL, density 1 g/mL, 400 g, 3 sig figs? Wait, 200 + 200 = 400 mL (3 sig figs, since 2.00×10² is 3). So m = 400 g (3 sig figs). c = 4.184 J/(g·°C) (4 sig figs, but we can keep it).

Then \(\Delta T = \frac{q}{mc} = \frac{9688.88}{400\times4.184} = \frac{9688.88}{1673.6} \approx 5.79^\circ C\).

T_f = 21.62 + 5.79 = 27.41, but check sig figs: initial temp 21.62 (4), but moles: 0.1724 (from 2.00×10² (3), 0.862 (3), 0.431 (3)). So heat: 56.2 (3) * 0.1724 (4) = 9.688 (4), but 56.2 is 3, so 9.69 kJ (3 sig figs). Then \(\Delta T = 9690 J / (400 g * 4.184 J/(g·°C)) = 9690 / 1673.6 ≈ 5.80 °C\). Then T_f = 21.62 + 5.80 = 27.42? Wait, maybe my intermediate steps had more sig figs, but let's recalculate:

Wait, 0.862 M * 0.200 L = 0.1724 mol (4 sig figs, since 0.200 is 3? Wait, 2.00×10² mL is 3 sig figs, so 0.200 L is 3. 0.862 is 3. So 0.862 * 0.200 = 0.1724, but we should round to 0.172 mol (3 sig figs). Then heat: 56.2 kJ/mol * 0.172 mol = 9.6664 kJ = 9666.4 J. Then \(\Delta T = 9666.4 / (400 * 4.184) = 9666.4 / 1673.6 ≈ 5.776 °C\). Then T_f = 21.62 + 5.776 ≈ 27.396 ≈ 27.4 °C? Wait, no, initial temp is 21.62 (4 sig figs), but the least number of sig figs in concentration is 3 (0.862, 0.431), volume is 3 (2.00×10²). So the final answer should have 3 sig figs? Wait, 21.62 is 4, but when adding, the decimal places: 21.62 (two decimal places) + 5.79 (two decimal places) = 27.41, which can be 27.4 °C? Wait, no, let's check the calculation again with precise numbers:

moles of H⁺: 0.862 * 0.200 = 0.1724 (exact, since 2.00×10² is 3 sig figs, but 0.200 is 3, 0.862 is 3, so 0.1724 is 4 digits, but 3 sig figs: 0.172).

Heat: 56.2 * 0.172 = 9.6664 kJ = 9666.4 J.

Mass: 400 g (exact, since 200 + 200 = 400, density 1 g/mL).

c = 4.184 J/(g·°C).

\(\Delta T = 9666.4 / (400 * 4.184) = 9666.4 / 1673.6 = 5.776 °C\).

T_f = 21.62 + 5.776 = 27.396 ≈ 27.4 °C? Wait, but maybe the initial calculation with 0.1724 mol is correct (since 2.00×10² is 3, but 0.200 is 3, 0.862 is 3, so 0.1724 is 4, but when multiplying by 56.2 (3 sig figs), the heat is 56.2 * 0.1724 = 9.68888 kJ (which is 9.69 kJ with 3 sig figs). Then \(\Delta T = 9690 / (400 * 4.184) = 9690 / 1673.6 ≈ 5.80 °C\). Then T_f = 21.62 + 5.80 = 27.42 ≈ 27.4 °C? Wait, I think the correct answer is approximately 27.4 °C, but let's check with the exact calculation:

q = 56.2 kJ/mol * 0.1724 mol = 9.68888 kJ = 9688.88 J.

m = 400 g, c = 4.184 J/(g·°C).

\(\Delta T = 9688.88 / (400 * 4.184) = 9688.88 / 1673.6 = 5.790 °C\).

T_f = 21.62 + 5.790 = 27.41 °C, which rounds to 27.4 °C (3 sig figs) or 27.41 (4). But the heat of neutralization is -56.2 (3 sig figs), concentration 0.862 (3), 0.431 (3), volume 2.00×10² (3). So the final answer should have 3 sig figs. So 27.4 °C? Wait, no, 21.62 is 4, but when adding, the number of decimal places: 21.62 has two decimal places, 5.79 has two, so the sum has two decimal places? Wait, 21.62 + 5.79 = 27.41, which is two decimal places. But let's see the problem: initial temp 21.62 (4 sig figs), heat of neutralization -56.2 (3), concentrations 0.862 (3), 0.431 (3), volume 2.00×10² (3). So the limiting is 3 sig figs? Wait, 2.00×10² is 3, 0.862 is 3, 0.431 is 3, 56.2 is 3. So the final answer should have 3 sig figs. 27.4 °C (3 sig figs) or 27.41 (4). Wait, maybe the correct answer is 27.4 °C, but let's check with the precise calculation:

After calculating, the final temperature is approximately 27.4 °C (or 27.41, but with 3 sig figs, 27.4). Wait, no, let's do the calculation again:

1. Moles of H⁺: 0.862 M * 0.200 L = 0.1724 mol
2. Moles of OH⁻: 2 * 0.431 M * 0.200 L = 0.1724 mol
3. Moles of reaction: 0.1724 mol
4. Heat released: 0.1724 mol * 56.2 kJ/mol = 9.68888 kJ = 9688.88 J
5. Mass of solution: 400 g (200 + 200 mL, density 1 g/mL)
6. \(\Delta T = q / (mc) = 9688.88 J / (400 g * 4.184 J/(g·°C)) = 9688.88 / 1673.6 ≈ 5.790 °C\)
7. Final temperature: 21.62 °C + 5.790 °C = 27.41 °C

Since the heat of neutralization has 3 significant figures, and the concentrations/volumes have 3, the final answer should have 3 significant figures. 27.4 °C (3 sig figs) or 27.41 (4). Wait, 27.41 rounded to 3 sig figs is 27.4. So the final answer is 27.4 °C? Wait, no, 27.41 is 4 sig figs, but with 3, it's 27.4. But let's check the problem statement: "Be sure your answer has the correct number of significant digits." The initial temperature is 21.62 (4), heat of neutralization -56.2 (3), concentrations 0.862 (3), 0.431 (3), volume 2.00×10² (3). The least number of significant digits in the given data (excluding initial temp) is 3, so the final answer should have 3 significant digits. 27.41 rounded to 3 is 27.4. So the final temperature is approximately 27.4 °C.

Answer:

\boxed{27.4}