Question

Under the same conditions, how many mL of 1 M KOH and 0.5 M $H_{2}SO_{4}$ solutions, respectively, when mixed to form at a total volume of 100 mL produces the highest rise in temperature?

• A. 67, 33
• B. 33, 67
• C. 40, 60
• D. 50, 50

Answer 1

Answer: (D)

$\underset{2 \, m o l}{2 K O H \left(\right. a q \left.\right)}+\underset{1 \, m o l}{H_{2} S O_{4} \left(\right. a q \left.\right)} \rightarrow K_{2}SO_{4}+H_{2}O+Heat$
Thus, maximum heat is liberated if 2 mol of KOH reacts with 1 mol $H_{2}SO_{4}$ (or if the ratio of the amounts of KOH to $H_{2}SO_{4}$ is 2 : 1).
Let x = volume of KOH, then 100 mL - x = volume of $H_{2}SO_{4}$
Number of mmol KOH $=x\times 1$ mmol of KOH
Number of mmol of $H_{2}SO_{4}=\left(\right.100-x\left.\right)\times 0.5$ mmol of $H_{2}SO_{4}$
Maximum heat is liberated if
$\frac{x \, m \, m o l \, o f \, K O H}{\left(\right. 100 - x \left.\right) \times 0.5 \, m \, m o l \, H_{2} S O_{4}}=\frac{2}{1}$
So $\text{x} \, \text{=} \, \text{50} \, \text{mL}$