A10.0ml sample of calcium hydroxide solution requirewd 26.85ml of 0.225 m hydrochoric acid for neutralization. calculate the nolar concentration of the bsae

The balanced equation for the acid base reaction is as follows
Ca(OH)₂ + 2HCl ---> CaCl₂ + 2H₂O
stoichiometry fo Ca(OH)₂ to HCl is 1:2
number of HCl moles reacted - concentration of HCl x volume 
number of HCl moles - 0.225 mol/L x 26.85 x 10⁻³ L = 6.041 x 10⁻³ mol
according to molar ratio 
2 mol of HCl reacts with 1 mol of Ca(OH)₂
therefore 6.041 x 10⁻³ mol of HCl reacts with - 6.041 x 10⁻³ /2 mol = 3.021 x 10⁻³ mol
number of Ca(OH)₂ moles in 10.0 mL -  3.021 x 10⁻³ mol 
therefore Ca(OH)₂ moles in 1000 mL - 3.021 x 10⁻³ mol / (10.0 x 10⁻³ L) = 0.302 mol/L
molarity of Ca(OH)₂ - 0.302 M

The answer is c fish

If it releases 5 tons a day and there are 7 days a take that answer and multiply by 5 percent (0.05) to find the amount released in one week. leave a comment if you need more .