26. The heat of neutralization of HCl(aq) by NaOH(aq) is produced. If 50.00 mL of 1.05 M NaOH is added to 25.00 mL of 1.86 M HCl, with both solutions originally at what will be the final solution temperature? (Assume that no heat is lost to the surrounding air and that the solution produced in the neutralization reaction has a density of and a specific heat capacity of) Petrucci, Ralph H.. General Chemistry (p. 291). Pearson Education. Kindle Edition.

Needed in problem text is the temperature of the acid and base solutions before reaction and the accepted (published) molar heat of neutralization (strong acids) for HCl by NaOH. => Assuming solution temperature is 25⁰C* for both acid and base solutions before reaction and the molar heat of neutralization for HCl by NaOH is 55.7Kj/mole,** then …

* Standard Thermodynamic conditions => 25⁰C (298K) & 1.00 Atm.

**(https://chemdemos.uoregon.edu/demos/Heat-of-Neutralization-HClaq-NaOHaq)

NaOH + HCl => NaCl + H₂O

=> 50ml(1.05M NaOH) + 25ml(1.86M HCl)

=> 0.05(1.05)mole NaOH + 0.025(1.86)mole HCl

=> 0.0525mole NaOH + 0.0465mole HCl

=> (0.0525 – 0.0465)mole NaOH excess + 0.0465mole NaCl + H₂O + Heat

=> 0.0060mole NaOH in excess + 0.0465mole NaCl + H₂O + Heat

Note => NaOH neutralizes 0.0465mole HCl (Limiting Reactant) and produces 0.0465mole NaCl & H₂O + Heat of Neutralization.

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Heat flow (Q) = Heat received by solvent water from the NaOH + HCl reaction

=> Q = mcΔT = mc(T₂ - T₁) = specific heat produced by 0.0465mole HCl

=> Q(m) = Molar Heat of Neutralization = Q/mole = mcΔT/n

- m = mass of solvent water receiving heat = (50ml + 25ml)1g/ml = 75g

- c = specific heat of water = 4.184j/g⁰C

- T₂ - T₁ = T₂ - 25⁰C

- Q(m) = 55,700 joules/mole (published heat of neutralization)

- n = moles of HCl neutralized = 0.0465mole HCl

=>Q(m) = mcΔT/n = 55,700j/mole = (75g)(4.184j/g⁰C)(T₂ - 25⁰C) /0.0465mole

Solving for T₂ => T₂ = 33.2⁰C