when 37 g of caoh is aded to 1 litre of 0.55m h2so4 the poh is
You need to find the caps key on the keyboard and use it. Did you know there is a difference between 0.55 m and 0.55 M? Which do you want. By the way, caoh doesn't exist.
Ca(OH)2 and 0.55 molar
To find the pOH in this scenario, we need to first calculate the concentration of the OH- ion after the reaction between Ca(OH)2 and H2SO4 is complete. Here's how you can calculate it:
1. Determine the number of moles of Ca(OH)2 present:
- Mass of Ca(OH)2 = 37 g
- Molar mass of Ca(OH)2 = 40.08 g/mol (Ca: 40.08 g/mol, O: 16.00 g/mol, H: 1.01 g/mol)
- Moles of Ca(OH)2 = mass / molar mass = 37 g / 40.08 g/mol
2. Convert the moles of Ca(OH)2 to moles of OH-:
- Each Ca(OH)2 molecule dissociates to produce 2 OH- ions, so moles of OH- = 2 * moles of Ca(OH)2
3. Calculate the concentration of OH-:
- Volume of H2SO4 solution = 1 L
- Concentration of OH- ion (in moles per liter) = moles of OH- / volume of solution
Now, to find the pOH, we need to use the concentration of OH-:
4. Calculate the pOH using the concentration of OH-:
- pOH = -log10[OH- concentration]
Note: Since we only have the initial concentration of H2SO4 given, we are assuming that the reaction is complete and Ca(OH)2 is fully consumed in this calculation.
Make sure to perform the calculations accurately to get the correct pOH value.