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Question Number 22879 by Tinkutara last updated on 23/Oct/17

The Enthalpy of neutralization of acetic acid and sodium hydroxide is −55.4 kJ. What is the enthalpy of ionisation of acetic acid?

$$\mathrm{The}\:\mathrm{Enthalpy}\:\mathrm{of}\:\mathrm{neutralization}\:\mathrm{of} \\ $$$$\mathrm{acetic}\:\mathrm{acid}\:\mathrm{and}\:\mathrm{sodium}\:\mathrm{hydroxide}\:\mathrm{is} \\ $$$$−\mathrm{55}.\mathrm{4}\:\mathrm{kJ}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{enthalpy}\:\mathrm{of} \\ $$$$\mathrm{ionisation}\:\mathrm{of}\:\mathrm{acetic}\:\mathrm{acid}? \\ $$

Answered by Physics lover last updated on 23/Oct/17

⇒ enthalpy of ionisation = (−55.4)−(−57.3) = 1.9 kj/eq

$$\Rightarrow\:{enthalpy}\:{of}\:{ionisation} \\ $$$$=\:\left(−\mathrm{55}.\mathrm{4}\right)−\left(−\mathrm{57}.\mathrm{3}\right) \\ $$$$=\:\mathrm{1}.\mathrm{9}\:{kj}/{eq} \\ $$

Commented by Tinkutara last updated on 23/Oct/17

Can you explain by writing chemical reactions?

$$\mathrm{Can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{by}\:\mathrm{writing}\:\mathrm{chemical} \\ $$$$\mathrm{reactions}? \\ $$

Commented by Physics lover last updated on 23/Oct/17

alright. CH_3 COOH ⇌ CH_3 COO^− +H^(+ ) ,ΔH=x ...i Na^+ + OH^(− ) + CH_3 COO^− +H^+ → Na^+ +CH_3 COO^− + H_2 O ,ΔH=−57.3 KJ ...ii i + ii CH_3 COOH + Na^+ +OH^− → CH_3 COONa +H_(2 ) O ΔH =− 55.4 KJ (given) using hesses law ⇒ −55.4 = −57.3 + x ⇒ x = 1.9 KJ alternate trick : just understand the concept the the amount of heat which has not been liberated has been used to ionize the acid.

$${alright}. \\ $$$$\:{CH}_{\mathrm{3}} {COOH}\:\rightleftharpoons\:{CH}_{\mathrm{3}} {COO}^{−} +{H}^{+\:} ,\Delta{H}={x}\:\:...{i} \\ $$$$ \\ $$$$ \\ $$$${Na}^{+} +\:{OH}^{−\:} +\:{CH}_{\mathrm{3}} {COO}^{−} +{H}^{+} \\ $$$$\rightarrow\:{Na}^{+} +{CH}_{\mathrm{3}} {COO}^{−} +\:{H}_{\mathrm{2}} {O}\:\:,\Delta{H}=−\mathrm{57}.\mathrm{3}\:{KJ}\:\:...{ii} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${i}\:+\:{ii} \\ $$$${CH}_{\mathrm{3}} {COOH}\:+\:{Na}^{+} +{OH}^{−} \:\rightarrow\:{CH}_{\mathrm{3}} {COONa} \\ $$$$+{H}_{\mathrm{2}\:} {O}\:\:\Delta{H}\:=−\:\mathrm{55}.\mathrm{4}\:{KJ}\:\left({given}\right) \\ $$$${using}\:{hesses}\:{law} \\ $$$$ \\ $$$$\Rightarrow\:−\mathrm{55}.\mathrm{4}\:=\:−\mathrm{57}.\mathrm{3}\:+\:{x} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{1}.\mathrm{9}\:{KJ} \\ $$$$ \\ $$$${alternate}\:{trick}\:: \\ $$$${just}\:{understand}\:{the}\:{concept}\:{the} \\ $$$${the}\:{amount}\:{of}\:{heat}\:{which}\:{has}\:{not} \\ $$$${been}\:{liberated}\:{has}\:{been}\:{used}\:{to} \\ $$$${ionize}\:{the}\:{acid}.\: \\ $$$$ \\ $$$$ \\ $$

Commented by Tinkutara last updated on 23/Oct/17

But only heat is 57.3 kJ when reaction of strong acid and strong base. So why why you have used 57.3 here?

$$\mathrm{But}\:\mathrm{only}\:\mathrm{heat}\:\mathrm{is}\:\mathrm{57}.\mathrm{3}\:\mathrm{kJ}\:\mathrm{when}\:\mathrm{reaction} \\ $$$$\mathrm{of}\:\mathrm{strong}\:\mathrm{acid}\:\mathrm{and}\:\mathrm{strong}\:\mathrm{base}.\:\mathrm{So}\:\mathrm{why} \\ $$$$\mathrm{why}\:\mathrm{you}\:\mathrm{have}\:\mathrm{used}\:\mathrm{57}.\mathrm{3}\:\mathrm{here}? \\ $$

Commented by Physics lover last updated on 23/Oct/17

well,actually that is the enthalpy of reaction of a acid{IONISED} and an IONISED base. no matter whether its strong or weak,both hav to be in ionic form.

$${well},{actually}\:{that}\:{is}\:{the}\:{enthalpy} \\ $$$${of}\:{reaction}\:{of}\:{a}\:{acid}\left\{{IONISED}\right\} \\ $$$${and}\:{an}\:{IONISED}\:{base}.\: \\ $$$${no}\:{matter}\:{whether}\:{its}\:{strong}\:{or} \\ $$$${weak},{both}\:{hav}\:{to}\:{be}\:{in}\:{ionic}\:{form}. \\ $$

Commented by Tinkutara last updated on 23/Oct/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Commented by Physics lover last updated on 23/Oct/17

you are welcome.And you can call me friend.

$${you}\:{are}\:{welcome}.{And}\:{you}\:{can} \\ $$$${call}\:{me}\:{friend}. \\ $$

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