a quantity of $2.00 imes10^{2}$ 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
$h^{+}(aq)+oh^{-}(aq) o h_{2}o(l)$
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.

Question

a quantity of $2.00	imes10^{2}$ 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
$h^{+}(aq)+oh^{-}(aq)	o h_{2}o(l)$
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.

Accepted Answer

# Explanation:
## Step1: Calculate moles of H⁺ and OH⁻
Volume of HCl = $2.00\times10^{2}\space mL = 0.200\space L$
Moles of H⁺ from HCl: $n_{H^+}=M\times V = 0.862\space mol/L\times0.200\space L = 0.1724\space mol$

Volume of $Ba(OH)_2$ = $0.200\space L$
Moles of $OH^-$ from $Ba(OH)_2$: $n_{OH^-}=2\times M\times V = 2\times0.431\space mol/L\times0.200\space L = 0.1724\space mol$

So, moles of $H_2O$ formed (from $H^+ + OH^-
ightarrow H_2O$) is $0.1724\space mol$ (since H⁺ and OH⁻ are in 1:1 ratio and both have 0.1724 mol).

## Step2: Calculate heat released (q)
Heat of neutralization ($\Delta H$) = $-56.2\space kJ/mol$ (negative because heat is released)
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
Assume density of solution is $1\space g/mL$ (since it's aqueous solution, density ≈ 1 g/mL)
Total volume of mixed solution = $0.200 + 0.200 = 0.400\space L = 400\space mL$
Mass of solution, $m = density\times volume = 1\space g/mL\times400\space mL = 400\space g$

## Step4: Calculate temperature change ($\Delta T$)
Specific heat capacity of water, $c = 4.184\space J/g^\circ C$
Using $q = mc\Delta T$, so $\Delta T=\frac{q}{mc}$
$\Delta T=\frac{9688.88\space J}{400\space g\times4.184\space J/g^\circ C}\approx5.74^\circ C$

## Step5: Calculate final temperature ($T_f$)
Initial temperature, $T_i = 21.62^\circ C$
$T_f = T_i + \Delta T = 21.62^\circ C + 5.74^\circ C = 27.36^\circ C$ (rounded to correct significant digits. Let's check significant digits:
- Molarity values: 0.862 (3 sig figs), 0.431 (3 sig figs), volume $2.00\times10^2$ mL (3 sig figs). So, the answer should have 3 decimal? Wait, no, let's recheck calculations. Wait, moles calculation: 0.862*0.200 = 0.1724 (4 sig figs from 0.862 and 0.200 (which is 3? Wait, 2.00×10² mL is 3 sig figs, so 0.200 L is 3 sig figs. 0.862 is 3, 0.431 is 3. So, moles of H⁺: 0.862*0.200 = 0.1724 (but 0.200 has 3, 0.862 has 3, so 3 sig figs? Wait, 0.862*0.200 = 0.1724, but we should consider significant digits. Wait, maybe my mistake earlier. Let's recalculate $\Delta T$ more accurately.

Wait, $q = 56.2\space kJ/mol\times0.1724\space mol$. 56.2 (3 sig figs), 0.1724 (4 sig figs from 0.862*0.200: 0.862 is 3, 0.200 is 3, so 0.862*0.200 = 0.1724 (but 0.200 is 3 sig figs, so 0.172 (3 sig figs)? Wait, no, 2.00×10² mL is 3 sig figs, so 0.200 L is 3 sig figs (the trailing zero after decimal is significant). So, 0.862 (3) * 0.200 (3) = 0.1724, but we take 3 sig figs: 0.172 mol? Wait, no, 2.00×10² is 3 sig figs, so 0.200 L is 3 sig figs. 0.862 is 3, 0.431 is 3. So, moles of H⁺: 0.862 * 0.200 = 0.1724 (but with 3 sig figs, it's 0.172 mol? Wait, no, 0.862 has 3, 0.200 has 3, so the product has 3 sig figs: 0.172 mol. Similarly, moles of OH⁻: 2*0.431*0.200 = 0.1724, 3 sig figs: 0.172 mol. Then, q = 56.2 * 0.172 = 9.6664 kJ = 9666.4 J. Then, $\Delta T = 9666.4 / (400 * 4.184) = 9666.4 / 1673.6 ≈ 5.776^\circ C$. Then, $T_f = 21.62 + 5.776 ≈ 27.396^\circ C$. Wait, maybe my initial sig fig handling was wrong. Let's check the problem:

Given: 2.00×10² mL (3 sig figs), 0.862 M (3), 0.431 M (3), 21.62 °C (4 sig figs), -56.2 kJ/mol (3). So, the limiting sig figs are 3 (from molarities and volume). Wait, but when we calculated moles, 0.862*0.200 = 0.1724 (but 0.200 is 3 sig figs, so 0.172 mol? No, 2.00×10² is 3 sig figs, so 0.200 L is 3 sig figs (the zero after decimal is significant, so 2.00×10² mL = 200. mL, which is 3 sig figs, so 0.200 L is 3 sig figs). So, 0.862 (3) * 0.200 (3) = 0.1724, but we should keep 3 sig figs: 0.172 mol? Wait, no, 0.862 * 0.200 = 0.1724, and since both have 3 sig figs, the result should have 3 sig figs: 0.172 mol. Similarly, 0.431*0.200*2 = 0.1724, 3 sig figs: 0.172 mol. Then, q = 56.2 * 0.172 = 9.6664 kJ = 9666.4 J. Then, m = 400 g (3 sig figs, since 400 mL with density 1 g/mL, 400 g is 3 sig figs? Wait, 400 mL: if it's 400. mL, it's 3 sig figs, but here it's 400 mL (from 200 + 200). So, 400 g (maybe 1 sig fig? No, no, the volumes are 2.00×10² mL (3 sig figs) each, so total volume 4.00×10² mL (3 sig figs), so mass 400. g (3 sig figs). So, m = 400. g (3 sig figs). c = 4.184 J/g°C (4 sig figs). So, $\Delta T = 9666.4 / (400. * 4.184) = 9666.4 / 1673.6 ≈ 5.776^\circ C$. Then, T_f = 21.62 + 5.776 ≈ 27.396, which rounds to 27.4 °C? Wait, no, let's do exact calculation:

Wait, let's recalculate moles correctly:

Volume of HCl: 2.00×10² mL = 0.200 L (3 sig figs)
Moles of HCl: 0.862 mol/L * 0.200 L = 0.1724 mol (so H⁺ is 0.1724 mol)

Volume of Ba(OH)₂: 0.200 L (3 sig figs)
Moles of Ba(OH)₂: 0.431 mol/L * 0.200 L = 0.0862 mol
Moles of OH⁻: 2 * 0.0862 mol = 0.1724 mol (so OH⁻ is 0.1724 mol)

So, moles of reaction (H⁺ + OH⁻) is 0.1724 mol (since both H⁺ and OH⁻ are 0.1724 mol, 1:1 ratio). So, that's correct.

Heat released: q = 56.2 kJ/mol * 0.1724 mol = 56.2 * 0.1724 = let's calculate 56 * 0.1724 = 9.6544, 0.2 * 0.1724 = 0.03448, total = 9.6544 + 0.03448 = 9.68888 kJ = 9688.88 J.

Mass of solution: 400 mL * 1 g/mL = 400 g (assuming density 1 g/mL, which is valid for dilute aqueous solutions).

Specific heat capacity of water: c = 4.184 J/g°C.

ΔT = q / (m * c) = 9688.88 J / (400 g * 4.184 J/g°C) = 9688.88 / (1673.6) ≈ 5.79 °C (wait, 400*4.184=1673.6, 9688.88 / 1673.6: 1673.6*5=8368, 9688.88-8368=1320.88, 1320.88/1673.6≈0.789, so total ΔT≈5.789≈5.79 °C)

Then, T_f = 21.62 + 5.79 = 27.41 °C. Rounding to correct significant digits:

Given data: 2.00×10² mL (3), 0.862 M (3), 0.431 M (3), 21.62 °C (4), -56.2 kJ/mol (3). The least number of sig figs in multiplication/division steps: 3 (from molarities and volume). But temperature change: q has 3 sig figs (56.2), m has 3 (400 g, if we consider 400 as 3 sig figs: 4.00×10²), c has 4. So, ΔT should have 3 sig figs? Wait, 56.2 (3) * 0.1724 (4) = 9.68888 (which we can keep as 9.69 kJ for 3 sig figs? No, 0.1724 is 4 sig figs, but 56.2 is 3, so q is 9.69 kJ (3 sig figs). Then, m is 400 g (3 sig figs), c is 4.184 (4). So, ΔT = 9690 J / (400 g * 4.184 J/g°C) = 9690 / 1673.6 ≈ 5.79 °C (3 sig figs? 9690 is 3 sig figs (9.69×10³), 400 is 3, 4.184 is 4. So, 9.69×10³ / (4.00×10² * 4.184) = (9.69 / (4.00 * 4.184)) × 10^(3-2) = (9.69 / 16.736) × 10 ≈ 0.579 × 10 = 5.79 °C (3 sig figs). Then, T_f = 21.62 + 5.79 = 27.41, which rounds to 27.4 °C? Wait, no, 21.62 is 4 sig figs, 5.79 is 3, so when adding, the number of decimal places: 21.62 has 2 decimal places, 5.79 has 2? No, 5.79 has two decimal places? No, 5.79 is two decimal places? Wait, 5.79 is two decimal places (the 9 is in hundredths place). 21.62 is two decimal places. So, when adding, the result should have two decimal places? Wait, no, significant digits for addition: the number of decimal places is determined by the least precise measurement. 21.62 has two decimal places, 5.79 has two decimal places, so T_f has two decimal places? Wait, 21.62 + 5.79 = 27.41, which is two decimal places. But let's check with exact calculation:

Wait, let's do all steps with exact values:

n = 0.1724 mol (exact from 0.862*0.200 and 2*0.431*0.200)

q = 56.2 * 0.1724 = 9.68888 kJ = 9688.88 J

m = 400 g (exact, since 200 + 200 mL, density 1 g/mL)

c = 4.184 J/g°C

ΔT = 9688.88 / (400 * 4.184) = 9688.88 / 1673.6 = 5.789 ≈ 5.79 °C

T_f = 21.62 + 5.79 = 27.41 °C. Rounding to the correct number of significant digits:

Looking at the given data:

- Volume: 2.00×10² mL (3 sig figs)
- Molarity: 0.862 M (3), 0.431 M (3)
- Initial temperature: 21.62 °C (4 sig figs)
- Heat of neutralization: -56.2 kJ/mol (3 sig figs)

The heat of neutralization has 3 sig figs, molarities and volume have 3, so the final answer should have 3 sig figs? Wait, 27.41 rounded to 3 sig figs is 27.4 °C? Wait, no, 27.41: first three sig figs are 2,7,4, the next digit is 1, which is less than 5, so 27.4 °C. But let's check with another approach.

Wait, maybe I made a mistake in moles. Wait, HCl is 0.862 M, 200 mL: moles of HCl = 0.862 * 0.2 = 0.1724 mol. Ba(OH)₂ is 0.431 M, 200 mL: moles of Ba(OH)₂ = 0.431 * 0.2 = 0.0862 mol. Each Ba(OH)₂ gives 2 OH⁻, so moles of OH⁻ = 2 * 0.0862 = 0.1724 mol

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Calculate moles of H⁺ and OH⁻

Step2: Calculate heat released (q)

Step3: Calculate total mass of solution

Step4: Calculate temperature change (\(\Delta T\))

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

Answer:

Step1: Calculate moles of H⁺ and OH⁻

Step2: Calculate heat released (q)

Step3: Calculate total mass of solution

Step4: Calculate temperature change (\(\Delta T\))

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