Question

Predict how the temperature change that you measure for 10 mL of 1 M NaOH added to 10 mL of 1 M hydrochloric acid compare to the temperature change when 20 mL of 1 M NaOH is added to 20 mL of 1 M hydrochloric acid.

Answer 1

Answer

Both temperature change would be equal

Explanation

NaOH + HCl ------> NaCl + H2O

Heat of neutralization of strong acid vs strong base is -57kJ /mol H2O

i) 10ml /10ml

number of moles of NaOH = ( 1mol/1000ml) ×10ml = 0.01mol

number of moles of HCl = (1mol/1000ml) × 10ml = 0.01mol

number of moles of H2O formed = 0.01mol

∆H = ( -57000J/1mol)× 0.01mol = -570J

Heat absorbed by the solution

q = m × ∆T × C_s

570J = 20g × ∆T × 4.184J/g℃

∆T = 6.81℃

ii) 20ml/20ml

number of moles of NaOH = (1mol/1000ml) × 20ml = 0.020mol

number of moles of HCl = (1mol/1000ml) × 20ml = 0.020mol

numbet of moles of H2O formed = 0.020mol

∆H =( -57000J/1mol) ×0.020mol = - 1140J

q = m × ∆T × C_s

1140J = 40g × ∆T × 4.184J/g℃

∆T = 6.81℃