Question

B. ΔH_solution for neutralization of HCl_(aq) and NaOH_(aq): CALCULATE DELTA H FOR NEUTRALIZATION
Volume of HCl: 0.0500 L
Volume of NaOH: 0.0500 L
Volume total: 0.100 L
Molarity of HCl: 1.95 M
Molarity of NaOH: 1.95 M
ΔT for reaction B: 11.0°C

Answer 1

HCl + NaOH --> H2O + NaCl dH

Q = m*Cp*(Tf-Ti)

m = mass of water solution + mass of HCl + mass of NaOH

mass of water solution = 0.1 L --> 0.1 kg or 100 g

mass of HCl = 1.95M*0.05L = 0.0975 mol of HCl

MW o fHCL )= 36

mass of HCl = 0.0975*36 = 3.51 g of HCl

apply same logif for NaOH

mol NaOH = M*V = 0.05*1.95 = 0.0975 mol

MW of NAO H= 40 g/mol

mass = 40*0.0975 = 3.9 g

Total mass = 100+3.9+3.51= 107.41 g

Cp = 4.184 J/gC (use the one of water since 99% is water)

Q = m*cP*(Tf-Ti)

Q = 107.41*4.184*(11) = 4943.4 J

but you wante dH for neutralization so base this on the

0.0975 mol of HCl/naOH

Therefore

E = Q/mol = 4943.4 J / 0.0975 mol = 50701.93 J per mol or 50.7 kJ/mol