Question

Q1. Assume the solubility product of Mg(OH)₂ is 2.2 × 10⁻¹¹ at a certain temperature. What minimum OH⁻ concentration must be attained (for example, by adding NaOH) to decrease the Mg²⁺concentration in a solution of Mg(NO₃)₂ to less than 1.1 × 10⁻¹⁰ M

Q2. Calculate the molar solubility of AgBr in a 2.2 M NH₃ solution.

Q3.Calculate the molar solubility of AgBr in a 2.0 M NH₃ solution.

Q4. A 10.0−mL solution of 0.360 M NH₃ is titrated with a 0.120 M HCl solution. Calculate the pH after the following additions of the HCl solution (a)at 0 (b)at 10ml (C) 30ml (D) 40ml

Please help-! I'll make sure i rate it if i get the right answer (to any questions above) Thank you in advance !!!!

Answer 1

The solubility product of Mg(OH)₂ is 1.2 x 10¯¹¹. What minimum OH¯ concentration must be attained (for example, by adding NaOH) to decrease the Mg²⁺ concentration in a solution of Mg(NO₃)₂ to less than 1.1 x 10¯¹⁰ M?

Solution:

K_sp expression:

K_sp = [Mg²⁺] [OH¯]²

We set [Mg²⁺] = 1.1 x 10¯¹⁰ and [OH¯] = x. Substituting into the K_sp expression:

1.2 x 10¯¹¹ = (1.1 x 10¯¹⁰) (x)² x = 0.33 M

Any sodium hydroxide solution greater than 0.33 M will reduce the [Mg²⁺] to less than 1.1 x 10¯¹⁰ M.

What is the solubility, in moles per liter, of AgCl (K_sp = 1.77 x 10^-10) in 0.0300 M CaCl₂ solution?

Solution:

1) Concentration of chloride ion from calcium chloride:

0.0300 M x 2 = 0.0600 M

from here:

CaCl₂(s) ---> Ca²⁺(aq) + 2Cl¯(aq)

2) Calculate solubility of Ag⁺:

K_sp = [Ag⁺] [Cl¯]

1.77 x 10^-10 = (x) (0.0600)

x = 2.95 x 10^-9 M

Since there is a 1:1 ratio between the moles of aqueous silver ion and the moles of silver chloride that dissolved, 2.95 x 10^-9 M is the molar solubility of AgCl in 0.0300 M CaCl₂ solution.

3) A 10.0 mL solution of 0.300 M NH3 is titrated with a 0.100 M HCl solution. Calculate the pH after the additio?

NH3 + H2O --> NH4+ + OH-
HCl reacts with the OH-

0.01L x 0.3M NH3 = 0.003moles NH3 which produces 0.003moles OH-
0.02L x 0.1M HCl = 0.002moles H+ into the system to neutralize 0.002moles NH3
this leaves 0.001moles NH3 in 0.03L or 0.033M NH3

Kb NH3 = 1.8x10^-5 = [NH4+][OH-]/[NH3]
1.8x10^-5 = x^2 / 0.033 - x
5.4x10^-8 - 1.8x10^-5x = x^2
quadratic equation
x = [OH-] = 2.32x10^-4M
pH = 14 - pOH = 10.4