Can you please help me with 6 and 7? Thank you! Experiment 11: Enthalpy 119 5. A
ID: 588248 • Letter: C
Question
Can you please help me with 6 and 7? Thank you! Experiment 11: Enthalpy 119 5. A S00 ml. solution of 0.805 M acetic acid, a monoprotic acid, is completely neutralized by trosid addition of 23.55 mL of sodium hydroxide. What is the concentration of the initial sodium hydrokio solution? Please show your work. 2-23.55 1 .205) ( 50m) , "-[T709M 23. 55nL) 6. In the reaction in question 5, the initial temperature of each solution is 22.1'C. The temperature mixing is 29.4°C. How much heat (i work. n J) is liberated by this neutralization reaction? Please show your 7. What is the heat of neutralization expressed in terms of kJ/mol for this reaction? Please show your workExplanation / Answer
Total volume of solution = 50.0+ 23.55 = 73.55 ml
If the density of solution is as water then the mass of solution = 73.55 g
Heat of solution , Q = mc dT
m = mass of solution= 73.55 g
c, specific heat of solution =4.184 J / g-degree C
dt = temperature change, 29.4 C – 22.1 C. =7.30 C.
Q= 73.55 g *4.184 J / g-degree C*7.30 C
= 2246.45 J
=2.25 kJ
CH3COOH + NaOH = CH3COONa + H2O
Moles of acid = molarirty * volumein L
= 0.805 *50/1000
= 0.04025 Mole
Heat of neutralization per mole = total heat in KJ/ number of moles
= 2.25 KJ /0.04025 MOLE
= 55.9 KJ/ mole