Question

Three student chemists measured 50.0 mL of 1.00 M NaOH in separate styrofoam cups calorimeters. Brett added 50.0 mL of 1.10 M HCl to his solution of NaO; Dale added 45.5 mL of 1.10 M HCl (equal moles) to his NaOH solution. Lyndsay added 50.0 mL of 1.00 M HCl to her NaOH solution. Each student recorded the temperature change and calculated the enthalpy of neutraliation. Identify the student who observes a temperature change that will be different from that observed by the other two chemists. Explain why and how (higher or lower) the temperature will be different.

Answer 1

Addition of HCl in NaOH results into the formation of NaCl and H₂O. this is called neutralisation reaction.

HCl + NaOH NaCl + H₂O

Volume of NaOH taken is 50 ml in all cases

Molarity of NaOH is 1 M.

the number of moles of NaOH is 50x1 = 50 mili moles

in first case

Volume of HCl is 50 ml

molarity of HCl is 1.10 M

the number of moles of HCl is 50x1.10 = 55 mili moles

In second case

Volume of HCl is 45 ml

molarity of HCl is 1.10 M

the number of moles of HCl is 45 x1.10 = 50.5 mili moles

In third case

Volume of HCl is 50 ml

molarity of HCl is 1.0 M

the number of moles of HCl is 50x1.0 = 50 mili moles

Therefore, in first case temperature change is different, because number of moles of HCl added is higher in first case while equal in second and third case.

the change in temperature depends on the number of moles of HCl added.