When 5.0 mL of a 1.0 M HCl solution is mixed with 5.0 mL of a 0.1 M NaOH solution, temperature of solution is increased by `2^@C` predicted accurately from this observation?

Options

A. If 10 mL of same HCl is mixed with 10 mL of same NaOH, temperature rise will be `4^@C`
B. If 10 mL of 0.05 HCl is mixed with 10 mL of 0.05 M NaCl the temperature rise will be `2^@C`
C. If 5 mL of 0.1 M HCl is mixed with 5 mL of 0.1 M `NH_3` solution, the temperature rise will be less than `2^@C`
D. If 5 mL 0.1 M is mixed with 5 mL of 0.1 M NaOH, the temperature rise will be less than `2^@C`

Answer

C, D

Explanation

Heat involved during the neutralisation `=(1.0 xx 1 xx 10^(-3)) xx c xx 2 = 0.02 c`
0.02c heat is evolved due to the neutralization or formulation of `(0.1 xx 5) = 0.5 m` moles of `H_2O` .
(A) Heat evolved `= 0.04c = (20 xx 1 xx 10^(-3)) xx c xx Delta T implies Delta T implies DeltaT = 2^@C [0.1 xx 10 = 1` m moles of `H_2O` formed]
(B) No temperature rise as there is no neutralization
(C) M moles of formed but heat evolved will be less than 0.04c as is a weak base.
(D) Similar to (C). [`CH_3COOH` is a weak acid]