**Passage 2 (continued):** <br> **Q3:** The volume of required for only in mL solution is: — Redox Reactions and Volumetric Analysis Chemistry Question
Question
**Passage 2 (continued):** <br> **Q3:** The volume of $HCl$ required for $NaHCO_3$ only in $25$ mL solution is:
Answer: C
💡 Solution & Explanation
**Step 1:** $V_{NaOH} = 5$ mL, $V_{Na_2CO_3} = 30$ mL. <br>**Step 2:** $V_{mo} = V_{NaOH} + V_{Na_2CO_3} + V_{NaHCO_3} = 35$ mL. <br>**Step 3:** $5 + 30 + V_{NaHCO_3} = 35 \Rightarrow V_{NaHCO_3} = 0$ mL. The mixture does not contain original $NaHCO_3$.
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