You want to analyze a cadmium nitrate solution. What mass of NaoH s to precipeat

Question

You want to analyze a cadmium nitrate solution. What mass of NaoH s to precipeate the cd ions from 35.3 mL of o 493 M cd(Nos2 solution? needed Periodic Table Fundamental My Notes o Ask Your Te -4 points BLB13 4P081. (a) what volume of 0.115 M Haoa solution is needed to neutralize 50.00 mL of 0.0870 M NaOH? neutralize 2.83 g of MgCOH)2? needed to (b) what volume of o 128 M HCI is (e) if 25.7 mL of AgNO, is needed precipitate all the cr ions in a 78s-mg sample of Kc (forming Agol), what is the molarity of the AgNO3 solution? to

Explanation / Answer

1) moles Cd2+ = 0.0353 L x 0.540 M=0.0190

the net ionic equation is
Cd2+ + 2 OH- = Cd(OH)2

moles OH- = moles NaOH = 2 x 0.0190 =0.0381

mass NaOH = 0.0381 mol x 40 g/mol=1.524g

2)HClO4 + NaOH ? NaClO4 + H2O

(50.00 mL) x (0.0870 M NaOH) x (1 mol HClO4 / 1 mol NaOH) / (0.115 M HClO4) = 37.826 mL HClO4

3)Mg(OH)2 + 2HCl --------> MgCl2 + 2H2O

First work out the number of moles of Mg(OH)2 you are starting with.

Moles = mass / molecular weight
molecular weight Mg(OH)2 = 24.31 + (2 x 16.00) + (2 x 1.008) = 58.326 g/mol
So moles Mg(OH)2 = 2.83 g / 58.326 g/mol
= 0.0485 moles of Mg(OH)2

Now, from the balanced equation you see that 2 moles of HCl react with 1 moles of Mg(OH)2. Therefore you need 2 x the number of moles of HCl as you have Mg(OH)2 to neutralise the solution.

moles HCl required = 2 x 0.0485 moles
= 0.09704 moles of HCl.

Now simply work out how many ml of 0.128 M HCl has this number of moles.

Molarity = moles / Litres
therefore litres = moles / Molarity
= 0.1008 moles / 0.128 M
= 0.758 litres
= 758 mL

4)AgNO3 + KCl => AgCl + KNO3

Mass of KCl = 785 mg = 0.785 g

Moles of AgNO3 = moles of KCl

= mass of KCl/molar mass of KCl

= 0.785/74.55 = 0.01053 mol

Volume of AgNO3 = 25.7 mL = 0.025.7L

Molarity of AgNO3

= moles of AgNO3/volume of AgNO3

= 0.01053/0.0257 = 0.409 M

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