Question

A 0.00100 mol sample of ca(oh)2 requires 25.00 ml of aqueous hcl for neutralization according to the reaction below. what is the concentration of the hcl? ca(oh)2(s) + 2hcl(aq) --> cacl2(aq) + h2o(l)

Answer 1

Answer:

0.08 mol/L is the concentration of the HCl solution.

Explanation:

$Ca(OH)_2(s) + 2HCl(aq)\rightarrow CaCl_2(aq) + H_2O(l)$

Moles of calcium = 0.00100 mol

According to reaction ,1 mole of calcium hydroxide reacts with 2 moles of HCl.

Then 0.00100 mole of calcium hydroxide will react with:

$\frac{2}{1}\times 0.00100 mol=0.00200 mol$

Moles of HCl = 0.00200 mol

Volume of solution = 25.00 mL = 0.025 L

$Concentration = \frac{Moles}{Volume(L)}$

Concentration of HCl solution:

$[HCl]=\frac{0.00200 mol}{0.025 L}=0.08 mol/L$

Answer 2

Ca(OH)₂ + 2HCl ---> CaCl₂ + 2H₂O
1 mol.........2 mol

1 mol Ca(OH)₂ --- 2 mol HCl
0,001 mol --------- X
X = 0,001×2 = 0,002 mol HCl

25 ml = 0,025 l = 0,025 dm³

C = n/V
C = 0,002/0,025
C = 0,08 mol/dm³