Answer:
0.136g
Explanation:
A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?
Initial mole of Co(NO3)2
Mole of Co(NO3)2 in final solution
Mole of NO3- in final solution = 2 x Mole of Co(NO3)2
Mass of NO3- in final solution is mole x Molar mass of NO3
This is an incomplete question, here is a complete question.
A 0.130 mole quantity of NiCl₂ is added to a liter of 1.20 M NH₃ solution. What is the concentration of Ni²⁺ ions at equilibrium? Assume the formation constant of Ni(NH₃)₆²⁺ is 5.5 × 10⁸
Answer : The concentration of ions at equilibrium is,
Explanation : Given,
Moles of = 0.130 mol
Volume of solution = 1 L
Concentration of = Concentration of
= 0.130 M
Concentration of = 1.20 M
The equilibrium reaction will be:
Initial conc. 0.130 1.20 0
At eqm. x [1.20-6(0.130)] 0.130
= 0.42
The expression for equilibrium constant is:
Now put all the given values in this expression, we get:
Thus, the concentration of ions at equilibrium is,