The heat of neutralization of (i) CHCl_2 – COOH by NaOH is 12830 cal, (ii) HCl by NaOH is 13680 cal and (iii) NH_4OH by HCl is 12270 cal. What is the heat of neutralization of dichloro acetic acid by NH_4OH? Calculate the heats of ionization of dichloro acetic acid and NH_4OH.

The heat of neutralization of (i) CHCl_2 – COOH by NaOH is 12830 cal

1.	The heat of neutralization of (i) CHCl_2 – COOH by NaOH is 12830 cal, (ii) HCl by NaOH is 13680 cal and (iii) NH_4OH by HCl is 12270 cal. What is the heat of neutralization of dichloro acetic acid by NH_4OH? Calculate the heats of ionization of dichloro acetic acid and NH_4OH.
Answer» Solution :(i) `CHCl_2COOH + OH^(-) to CHCl_(2)COO^(-) + H_2O, ""DeltaH_(1)= 12830 cal` (ii) `H^(+) + OH^(-) to H_2O, ""DeltaH_(2) = 13680 cal` (iii) `NH_4OH + H^(+) to NH_(4)^(+) + H_2O, ""DeltaH_3 = -12270 cal` Consider (iv) `CHCl_(2)COOH + NH_4OH to CHCl_2COO^(-) + NH_(4)^(+) + H_2O` (v) Will be obtained by APPLYING `[(i) + (iii) - (ii)]` `Delta H (iv) = Delta H_(1) + Delta H_(3) - DeltaH_(2) = -12830 - 12270 + 13680 = -11420 cal` Now, (vi) `CHCl_(2)COOH to CHCl_(2)COO^(-) + H^(+)` (VII) Will be obtained by applying (i) - (ii) `:. Delta H(v) = DeltaH_(1) - DeltaH_2 = 13680 - 12830 = 850 cal//mol` Similarly `DeltaH_("ionization")` for `NH_(4)OH = 13680 - 12270 = 1410 cal// mol`.