(JEE Main 2021) of is added to of . The molarity of in final solution is: — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
(JEE Main 2021) $20 \text{ mL}$ of $0.1 \text{ M } H_2SO_4$ is added to $30 \text{ mL}$ of $0.2 \text{ M } KOH$. The molarity of $OH^-$ in final solution is:
Answer: A
💡 Solution & Explanation
$\text{mEq Acid} = 20 \times 0.1 \times 2 = 4$. <br>$\text{mEq Base} = 30 \times 0.2 \times 1 = 6$. <br>$\text{Excess Base} = 2 \text{ mEq} = 2 \text{ mmol } OH^-$. <br>$\text{Molarity} = 2 / 50 = 0.04 \text{ M}$.
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