**Passage 2:** g of a mixture of and is dissolved in water to make 1 L solution. mL of this solution — Redox Reactions and Volumetric Analysis Chemistry Question
Question
**Passage 2:** $10$ g of a mixture of $NaOH, Na_2CO_3$ and $NaHCO_3$ is dissolved in water to make 1 L solution. $25$ mL of this solution required $20$ mL of $0.1 M$ $HCl$ with phenolphthalein. Another $25$ mL of the same solution required $35$ mL of $0.1 M$ $HCl$ with methyl orange. <br> **Q1:** The milliequivalents of $NaOH$ in $25$ mL of the solution is:
💡 Solution & Explanation
**Step 1:** $V_{ph} = V_{NaOH} + 1/2 V_{Na_2CO_3} = 20$ mL. <br>**Step 2:** $V_{mo} = V_{NaOH} + V_{Na_2CO_3} + V_{NaHCO_3} = 35$ mL. <br>**Step 3:** Volume for $1/2 Na_2CO_3$ (from Ph to MO) = $V_{mo} - V_{ph} = 35 - 20 = 15$ mL. <br>**Step 4:** Thus, full $Na_2CO_3$ requires $2 \times 15 = 30$ mL. <br>**Step 5:** Since $V_{NaOH} + 15 = 20$, then $V_{NaOH} = 5$ mL. <br>**Step 6:** Meq $NaOH = 5 \times 0.1 = 0.5$.