Question

Data Table A. Stock solution concentrations of HCl, H₃PO₄ and NaOH

[HCl]	1.5 M
[H3PO4]	1.0 M
[NaOH]	1.5 M

Data Table B. Temperature data for combinations of NaOH and HCl(aq)

Expt #	mL NaOH	mmol NaOH	ml H2O	mL HCl	mmol HCl	Initial T, to the 0.01  °C	Final T, to the 0.01  °C	ΔT, °C
1	20.	30	20.	10.	15	25	29	4
2	20.	30	10.	20.	30	24.5	32	7.5
3	20.	30	0	30.	45	24.5	32	7.5

Data Table C. Temperature data for combinations of NaOH and H₃PO₄

Expt #	mL NaOH	mmol NaOH	ml H2O	mL H3PO4	mmol H3PO4	Initial T, to the 0.01 °C	Final T, to the 0.01 °C	ΔT,  °C
4	15.	23	30.	15.	15	22	26	4
5	30.	45	15.	15.	15	21	29	8
6	45.	68	0	15.	15	20.5	29	8.5

(a) Based on the amount of product formed in each of the mixing experiments (that is, the "final" line in each reaction table), which experiments involving NaOH and HCl, if any, would you expect to give the same temperature changes? (Select all that apply.)

experiment #1

experiment #2

experiment #3

(b) Based on the amount of product formed in each of the mixing experiments (that is, the "final" line in each reaction table), which experiments involving NaOH and H₃PO₄, if any, would you expect to give the same temperature changes? (Select all that apply.)

experiment #4

experiment #5

experiment #6

Answer 1

a)
temperature change will be same if amount of product formed is same

for experiment 1:
mmol of NaOH = 30
mmol of HCl = 15
HCl is limiting reagent and hence mmol of product formed = 15

for experiment 2:
mmol of NaOH = 30
mmol of HCl = 30
Both reactant are in equal proportion and hence mmol of product formed = 30

for experiment 3:
mmol of NaOH = 30
mmol of HCl = 45
NaOH is limiting reagent and hence mmol of product formed = 30

Answer:
experiment #2

experiment #3

b)
temperature change will be same if amount of product formed is same
H3POH has 3 H+

for experiment 4:
mmol of OH- = 23
mmol of H+ = 15*3 = 45
OH- is limiting reagent and hence mmol of product formed = 23

for experiment 5:
mmol of OH- = 45
mmol of H+ = 15*3 = 45
mmol of product formed = 45

for experiment 6:
mmol of OH- = 68
mmol of H+ = 15*3 = 45
H+- is limiting reagent and hence mmol of product formed = 45

Answer:
experiment #5

experiment #6