Question

WILL MARK BRAINLIEST!!! 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

Answer 1

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.

Answer 2

Answer: The concentration of $Ca(OH)_2$ comes out to be 0.111 M.

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.185M\\V_1=32.4mL\\n_2=2\\M_2=?M\\V_2=27mL$

Putting values in above equation, we get:

$1\times 0.185\times 32.4=2\times M_2\times 27\\\\M_2=0.111M$

Hence, the concentration of $Ca(OH)_2$ comes out to be 0.111 M.