Question
Water is accidentally added to $350.00 \mathrm{~mL}$ of a stock solution of $6.00 \mathrm{M} \mathrm{HCl}$. A 75.00-mL sample of the diluted solution is titrated to $\mathrm{pH} 7.00$ with $78.8 \mathrm{~mL}$ of $4.85 \mathrm{M} \mathrm{NaOH}$. How much water was accidentally added? (Assume that volumes are additive.)
Step 1
We can do this by multiplying the molarity of the $\mathrm{NaOH}$ solution by its volume (in liters). This gives us: \[4.85 \mathrm{M} \times 0.0788 \mathrm{L} = 0.382 \mathrm{moles} \, \mathrm{of} \, \mathrm{NaOH}\] Show more…
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Water is accidentally added to $350.00 \mathrm{~mL}$ of a stock solution of $6.00 \mathrm{M}$ HCl. A $75.00-\mathrm{mL}$ sample of the diluted solution is titrated to $\mathrm{pH} 7.00$ with $78.8 \mathrm{~mL}$ of $4.85 \mathrm{M}$ $\mathrm{NaOH} .$ How much water was accidentally added? (Assume that volumes are additive.)
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