Question

What volume, in millilitres, of 0.510M Ca(OH)2 is needed to completely neutralise 231 mL of a 0.200 MHl solution? ​

Answer 1

To completely neutralize 231 mL of 0.200 MHl solution, you would need 45.3 mL of 0.510M Ca(OH)2.

Step-by-step explanation:

To determine the volume of 0.510M Ca(OH)2 needed to neutralize 231 mL of 0.200 MHl solution, we can use the balanced chemical equation:

HBr(aq) + Ca(OH)2(aq) → CaBr2(aq) + H2O(l)

From the equation, we can see that 2 moles of HBr reacts with 1 mole of Ca(OH)2. Therefore, we can use the mole-mole ratio to calculate the number of moles of Ca(OH)2 in 231 mL of 0.200 MHl solution:

1. Convert the volume of the Hl solution to moles: 231 mL * 0.200 M = 46.2 mmol Hl
2. Use the mole-mole ratio to find the moles of Ca(OH)2: 46.2 mmol Hl * (1 mol Ca(OH)2 / 2 mol HBr) = 23.1 mmol Ca(OH)2
3. Finally, calculate the volume of 0.510M Ca(OH)2 needed to neutralize 23.1 mmol Ca(OH)2: volume = moles / concentration = 23.1 mmol / 0.510 M = 45.3 mL