Question

The neutralization of HPO with KOH is exothermic 4 Н, РО, (аq) + 3 КОНаq) 3 H2O(l) + K3PO4 (aq) 173.2 kJ If 65.0 mL of 0.220 M H PO4 is mixed with 65.0 mL of 0.660 M KOH initially at 21.49 °C, predict the final temperature of the solution, assuming its density is 1.13 g/mL and its specific heat is 3.78 J/(g.°C). Assume that the total volume is the sum of the individual volumes 333.4 °C Trinal

Answer 1

Total volume = 65 + 65 = 130 mL, density = 1.13 g/mL

mass (m) = 130 mL x 1.13 g/mL = 146.9 g

moles H₃PO₄ = 65 mL x 10^-3 L x 0.220 M = 0.0143 mol

moles KOH = 65 mL x 10^-3 L x 0.660 M = 0.0429 mol

1 mol acid requires 3 mol base to neutralize

0.0143 x 3 = 0.0429, exact molar ratioo

energy released (q) = 0.0143 mol x 173.2 kJ/mol = 2.47676 kJ = 2476.76 J

specific heat (C_s) = 3.78 J/g.⁰C

q = m x C_s x ΔT

ΔT = q / (m x C_s )

ΔT = T_final - T_initial = 2476.76 J / (146.9 g x 3.78 J/g.⁰C) = 4.46 ⁰C

T_final - T_initial = 4.46 ⁰C

T_final = 4.46 ⁰C + T_initial = 4.46 ⁰C + 21.49 ⁰C = 25.95 ⁰C

T_final = 25.95 ⁰C

The neutralization of HPO with KOH is exothermic 4 Н, РО, (аq) + 3 КОНаq) 3 H2O(l) + K3PO4 (aq) 173.2 kJ If 65.0 mL of 0.220 M H3PO4 is mixed with 65.0 mL of 0.660 M KOH initially at 21.49 °C, predict the final temperature of the solution, assuming its density is 1.13 g/mL and its specific heat is 3.78 J/(g.°C). Assume that the total volume is the sum of the individual volumes 25.95 °C Trinal

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