Whenever an acid is neutralised by a base, the net reaction is
`H^(+) (aq) + OH^(-) (aq) to H_2O(l) , Delta H = -57.1 kJ`
Calculate the heat evolved for the following experiments

Explanation

According to the reaction
` H^(+) (aq) + OH^(-) (aq) to H_2O (l) " " Delta H = -57.1 kJ`
i.e., when one mole of `H^+` ions reacts with one mole of `OH^-` ions, one mole of water is formed and 57 .1 . kJ of energy is released.
(i) 0.50 mole of HCl solution is neutralised by 0.50 mole of NaOH solution.
0.50 mole of HCl = 0.50 mole of`H^+` ions
0.50 mole of NaOH= 0.50 mole of `OH^-` ions
On mixing, 0.50 mole of` H_2O` is formed.
Heat evolved for the formation of 0.50 mole of water `= 57.1 xx 0.5 = 28.55 kJ`
( ii) 0.50 mole of `HNO_3` solution is neutralised by 0.30 mole of NaOH solution
0.50 mole of `HNO_3` = 0.50 ,mole of` H^+` ions
0.30 mole of KOH = 0.30 mole of `OH^-` ions
i.e., 0.30 mole of `H^+` ions react with 0.30 mole of `OH^-` ions and fonns 0.30 mole of water. Heat evolved in the formation of 0.30 mole of `H_2O = 57.1 xx 0.3 = 17.13 kJ`
(iii) 100 ml of 0.2 M HCl is mixed with 100 ml of 0.3M NaOH solution
100 ml of 0.2 M HCl will give `( (0.2 xx 100 )/(1000) ) ` mole of = 0.02 mole of `H^+` ions .
100ml of 0.3 M NaOH will give `( (0.3 xx 100)/(1000) )` mole of = 0.3 mole of `OH^(-)` ions .
i.e., 0.02 mole of `H^+` Ions will react 0.03 mole of `OH^-` ions and produce 0.02 mole of `H_2O` molecules.
Heat evolved in the formation of 0.02 mole of `H_2O = 0.02 xx 57.1 = 1.142 kJ`
(iv) 400ml of 0.2M `H_2SO_4` is mixed with 600ml of 0.1 M KOH solution .
400 ml of 0.2 M `H_2SO_4` will give ` ( (2 xx 0.2)/(1000) ) xx 400` = 0.16 mole of `H^+` ions
600 ml of 0.1 M KOH will give ` ( (0.1 xx 600 )/(1000) ) = 0.06` mole of `OH^(-)` ion
Heat evolved in the formation of 0.06 mole of `H_2O = 0.06 xx 57.1 = 3.426 kJ`