Question

44.00 ml of 0.50 m h₂so₄ reacts with 0.35 m lioh and reaches the endpoint using bromthymol blue as an indicator. what volume of lioh is required for this reaction?
2lioh + h₂so₄ → li₂so₄ + 2h₂o
? ml lioh

Explanation:

Step1: Find moles of \( H_2SO_4 \)

Molarity formula: \( M = \frac{n}{V} \), so \( n = M \times V \).
Volume of \( H_2SO_4 \) is \( 44.00 \, \text{mL} = 0.044 \, \text{L} \), molarity \( M = 0.50 \, \text{M} \).
\( n_{H_2SO_4} = 0.50 \, \text{mol/L} \times 0.044 \, \text{L} = 0.022 \, \text{mol} \).

Step2: Use stoichiometry to find moles of \( LiOH \)

From reaction: \( 2 \, \text{mol } LiOH \) reacts with \( 1 \, \text{mol } H_2SO_4 \).
So \( n_{LiOH} = 2 \times n_{H_2SO_4} = 2 \times 0.022 \, \text{mol} = 0.044 \, \text{mol} \).

Step3: Find volume of \( LiOH \)

Molarity of \( LiOH \) is \( 0.35 \, \text{M} \). Using \( V = \frac{n}{M} \):
\( V_{LiOH} = \frac{0.044 \, \text{mol}}{0.35 \, \text{mol/L}} \approx 0.1257 \, \text{L} = 125.7 \, \text{mL} \) (rounded appropriately).

Answer:

\( 126 \) (or \( 125.7 \)) mL (Note: Depending on rounding, 126 or 125.7 is acceptable. If precise, \( \approx 125.7 \) mL, or 126 mL when rounded to whole number.)