If 27.0 mL of Ca(OH)2 with an unknown concentration is neutralized by 32.40 mL of 0.185 M HCl, what is the concentration of the Ca(OH)2 solution?
2n Ca(OH)2= n HCl
2[c Ca(OH)2 v Ca(OH)2] = c HCl v HCl
c Ca(OH)2= c HCl v HCl/2v Ca(OH)2
c Ca(OH)2= (0.158M)(32.40mL)/2(27mL)
c Ca(OH)2= 0.111M
pls. check if it's right:)
To find the concentration of the Ca(OH)2 solution, we can use the concept of stoichiometry and neutralization reactions. In a neutralization reaction, the reactants combine in a 1:1 ratio. The balanced chemical equation for the reaction between Ca(OH)2 and HCl is:
Ca(OH)2 + 2HCl -> CaCl2 + 2H2O
From the equation, we can see that 1 mole of Ca(OH)2 reacts with 2 moles of HCl.
First, let's find the number of moles of HCl used in the reaction:
Moles HCl = Volume HCl (in L) x Concentration HCl (in M)
= 0.03240 L x 0.185 mol/L
= 0.005996 mol
Since the reaction has a 1:1 stoichiometric ratio with respect to Ca(OH)2 and HCl, the moles of Ca(OH)2 is also 0.005996 mol.
Now, let's find the concentration of the Ca(OH)2 solution:
Concentration Ca(OH)2 (in M) = Moles Ca(OH)2 / Volume Ca(OH)2 (in L)
= 0.005996 mol / 0.0270 L
= 0.2224 M
Therefore, the concentration of the Ca(OH)2 solution is 0.2224 M.