Question

The temperature of 5 mL of a strong acid is increased by $5^{\circ }C$ when 5 mL of a strong base is added to it. If 10 mL of each of the same solutions is mixed, the temperature should increase by:

• A. $5^{\circ }$
• B. ${10}^{\circ }$
• C. ${15}^{\circ }$
• D. Cannot be predicted

Answer 1

The correct option is A $5^{\circ }$

For a strong acid and a strong base, the enthalpy of neutralisation is -13.7 kcal when meq. of acid = meq of base
It means that if the number of meqv. is same the amount of energy released is the same and hence the temprature rise will be the same.
Here, in this question the concentration of acid and base is same for 5 mL solution and 10 mL solution which means the same energy will be evolved in both the cases and hence the temperature rise will be same.

Let the concentration of strong acid and strong base = $cM$

For the reaction $H^{+}+{OH}^{-}⟶H_{2}O$

This 1 mole of acid reacts with 1 mole of base to form 1 mole of water with the liberation of heat.

Thus on mixing 5 ml $(5c\times {10}^{-3})$ thus amount of heat generated will be = $H\times 5c\times {10}^{-3}J$

Heat generated = $m.C_p\Delta T$

Here m = mass of solution

C = specific heat of water

$\Delta T=temperaturechange$

Let us consider the density of solution be $\frac{dg}{ml}$

Hence 5 ml of each acid and base requires

$H\times 5c\times 10^{-3}J=d\times (5+5)\times c\times \Delta T_1$

$\Delta T_1=\frac{H\times 5c\times {10}^{-3}}{10cd}$

${\Delta T}_1=\frac{Hc}{2cd}\times {10}^{-3}$

$Similarlyfor10mleachacidandbasesolutionrequires$

$H\times 10cJ=d\times (10+10)\times c\times {\Delta T}_2$

${\Delta T}_2=\frac{10cH}{20cd}\times {10}^{-3}$

${\Delta T}_2=\frac{Hc}{2cd}\times {10}^{-3}$

$Thus\Delta T_1={\Delta T}_2$

$Thusthereactionofstrongacidwithstrongbaseisaneutrilisationreaction.Thetemperatureof5mlofstrongacidincreasesby{5}^{o}Cwhen5mlofstrongbaseisaddedtoit.$