A sample of 9.49 g of solid calcium hydroxide is added to 30.0 mL of 0.490 M aqueous hydrochloric acid. Write the balanced chemical equation for the reaction. Physical states are optional.

Ca(OH)2(s) + 2HCl(aq) →  CaCl2(aq) + 2H2O(l)

Explanation:

Step 1: Data given

Mass of Ca(OH)2 = 9.49 grams

Volume of a 0.490 M HCl solution = 30.0 mL = 0.030 L

Step 2: The unbalanced equation

Ca(OH)2 + HCl →  CaCl2 + H2O

On the left side we have 2x O (in Ca(OH)2 )  and on the right side we have 1x O (in H2O). To balance the amount of O we have to multiply H2O, on the right side , by 2.

Ca(OH)2 + HCl →  CaCl2 + 2H2O

On the right side we have 4x H (in 2H2O), and on the left side we have 3x H  (2x in Ca(OH)2 and 1x in HCl). To multiply the amount of H, we have to multiply HCl on the left side by 2.

Ca(OH)2(s) + 2HCl(aq) →  CaCl2(aq) + 2H2O(l)

the dissociation of nickel hydroxide is follows:

the concentration of hydroxide ion is

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

thus, concentration of ni is 0.23 * 10^-^4 mol /l