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Part A 2 5 C. The endpoint was reached when 75,00 mL of the manganese() A saturated solution of manganese Rydroxide was prepared and an acid base station was performed to determine is K hydroxide solution was tivated with 2.83 ml of 0.0020 M ICT solution. What is the K ofmanganese hydroxide? IVO AE ROB? 1.92.10 Submit Previous Answers Request Answer X Incorrect: Try Again; 2 attempts remaining

Answer 1

Ans :-

Reaction between base Mn(OH)₂ and HCl is :

Mn(OH)₂ + 2 HCl -----------------> MnCl₂ + 2 H₂O

We know,

Molarity = Number of moles of solute/ Volume of solution in L

So,

Number of moles of HCl = Molarity x Volume of solution in L

= 0.0020 M x 0.00283 L

= 5.66 x 10^-6 mol

Because, 1 mol of base Mn(OH)₂ reacts with 2 moles of acid HCl

So,

Moles of base Mn(OH)₂ = 5.66 x 10^-6 mol / 2 = 2.83 x 10^-6 mol

Given, volume of solution = 75.00 mL = 0.075 L

So,

Molarity of Mn(OH)₂ = Number of moles of Mn(OH)₂ / Volume of solution in L

= 2.83 x 10^-6 mol / 0.075 L

= 3.77 x 10^-5 M

So,

[Mn²⁺] = 3.77 x 10^-5 M

Also,

[OH^-] = 2 x 3.77 x 10^-5 M

= 7.54 x 10^-5 M

Partial dissociation of Mn(OH)₂ in aqueous solution will be :

Mn(OH)₂ (s) <--------------------> Mn²⁺ (aq) + 2 OH^- (aq)

Now,

Expression of solubility product (K_sp) ( Solubility product is equal to the product of the molar concentration of products raised to the power of stoichiometric coefficient with respect to each reagent at equilibrium stage of the reaction).

K_sp = [Mn⁺²][OH^-]²

Substitute the values of [Mn²⁺] and [OH^-]​​​​​​​ in this equation, we have

K_sp = ( 3.77 x 10^-5 M ).(7.54 x 10^-5 M)²

K_sp = 2.14 x 10^-13 M³

Hence, K_sp of Mn(OH)₂ = 2.14 x 10^-13