Will mark ! 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? show all of the work needed to solve this problem. ca(oh)2 + 2hcl yields 2h2 o + cacl2

Ca(OH)2(aq) + 2HCl(aq) --> CaCl2(aq) + 2H2O(l) 
n(Ca(OH)2) = ½ x n(HCl) = (½) (0.03240 L) (0.185 M) = 0.003 mol 
[Ca(OH)2] = 0.003 mol / 0.0270 L = 0.111 M

0.185 M HCL = each liter of HCL holds 0.185 moles. So, 1 mL contains 0.185 millimoles of HCL. 32.40 mL * 0.185 = 5.994 mmol of HCL. 

Divide 5.994 mmol of HCL by 2 to get the amount of Ca(OH)2, which turns out to be 2.997 mmol. Then divide by 27, to get the amount of Ca(OH)2 in mmol in 1 mL of solution. 2.997 mmol / 27 mL = 0.111 M. We will have a 0.111 M concentration.

Ur e n d is not real