Question

When $10ml$ of a strong acid is added to $10ml$ of an alkali the temperature rises by ${5^0}C$ . If $100ml$ of the same acid mixed with $100ml$ of the same base, the temperature would be:
\[
  A.{\text{ }}{5^0}C \\
  B.{\text{ }}{50^0}C \\
  C.{\text{ }}{20^0}C \\
  D.{\text{ Cannot be predicted}} \\
 \]

Answer 1

Hint- In order to deal with this question we will use the concept that when amount of strong acid and an alkali are reacting in different volume but in equal ratio, the rise in temperature will be the same so here in this question we will use the old data for 10 ml to find the temperature rise at 100 ml.

Complete answer:
As we know that when a strong acid and strong base of particular amounts react, neutralisation takes place and there is a specific enthalpy change of neutralisation. ( $\Delta H$ of neutralisation)
When $10ml$ of a strong acid is added to $10ml$ of an alkali the temperature rises by ${5^0}C$
In this case, it is $\Delta H = {5^0}C$ .
As we know that when the amount of strong acid and an alkali are reacting in different volumes but in equal ratio, the rise in temperature will be the same.
No matter how much acid or solid base you use, the increase in enthalpy remains the same as the increase in enthalpy from neutralization.
So, when $100ml$ of the same acid mixed with $100ml$ of the same base.
Temperature rise still remains the same $\Delta H = {5^0}C$ .
Hence, the temperature would be ${5^0}C$ .

So, the correct answer is option A.

Note- A neutralization reaction can be described as a chemical reaction in which the acid and base react together quantitatively to form salt and water as products. In a neutralization reaction, there is a combination of ${H^ + }$ ions and $O{H^ - }$ ions which form water. A double displacement reaction is a type of neutralization. A salt formed is always the product of an acid-base reaction.