calculate the total concentration, in molarity, of hydroxide ions when 1,186.8 milligrams of gold(iii) hydroxide are completely dissolved into 0.154 l of 1.76 x10-1 m naoh. assume the added solute does not contribute significantly to the total volume. Report your answer to three significant digits.

Molarity or molar concentration is the number of moles of solute per liter of solution or molarity can be defined to be as number of moles of solute in volume of solution.

Mass of Gold(III) hydroxide = 1.1868 g

Moles of gold(III) hydroxide = mass of Au(OH)₃ / Molar mass of Au(OH)₃

Moles of Gold(III) = 1.1868/247.98 = 0. 004 moles

Moles of hydroxide ion from Au(OH)₃ will be,

Au(OH)₃ ⇔ Au³⁺ + OH⁻

Here,

if 1 mole of Au(OH)₃ gives 3 moles of OH⁻ ion then,

0.006 moles of Au(OH)₃ will give,

= 3 x 0.006 moles of OH⁻ ions

= 0.018 moles of OH⁻ ions

Again,

NaOH ⇔Na⁺ + OH⁻

Since, [OH⁻] = [NaOH] = 1.76 x 10⁻¹ = 0.176 M = molarity.

Hence, the moles of OH⁻ = Molarity of OH⁻ x VH of OH⁻

= 0.176 M x 0.154L

= 0.0271 moles

Now, total moles of OH⁻ will be

= OH⁻ from Au(OH)₃ + OH⁻ from NaOH

= (0.018 + 0.0271) moles

= 0.0451 moles of OH⁻

To find the concentration of hydroxide ion, we need to just apply the formula of molarity:

[OH⁻] = moles of OH⁻ / Volume of the solution

= 0.0451 / 0.154

= 0.292 M

Therefore, the the total concentration of hydroxide ion, in molarity is 0.292 M.

To learn more about molarity, refer: https://brainly.com/question/16727614

#SPJ4