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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: dudebuddyguy on November 21, 2013, 03:50:21 PM

Title: Zinc sulfare + Sodium hydroxide questions
Post by: dudebuddyguy on November 21, 2013, 03:50:21 PM
Trying to figure out the total ionic equation and the number of spectator ions.

I've got the balanced chemical equation;
2 NaOH + ZnSO4 = Na2SO4 + Zn(OH)2

And the net ionic equation
Zn2+(aq) + 2 OH-(aq) -> Zn(OH)2(s)

But I can't figure out how to find the number of spectator ions, and I have no clue what the total ionic equation is.

Backstory: Doing College Prep Chemistry to get my grade 12 science for an apprenticeship. I did graduate high school, but I got my last 9 credits via a work program instead of actual school (kicking myself nowadays)
Title: Re: Zinc sulfare + Sodium hydroxide questions
Post by: Borek on November 21, 2013, 03:58:31 PM
In your first equation just write everything as dissociated - that would be total.

Then see what was ignored when getting to net ionic - these all were spectators.
Title: Re: Zinc sulfare + Sodium hydroxide questions
Post by: dudebuddyguy on November 21, 2013, 04:04:11 PM
Quote from: Borek on November 21, 2013, 03:58:31 PM
In your first equation just write everything as dissociated - that would be total.

Then see what was ignored when getting to net ionic - these all were spectators.

Dissociated....what (I have my grade 10 science, but I've never done anything with chemistry until now, so forgive me for my stupidity)

As for the spectators....from what I can see....the 2 bits of Sulfur?

EDIT: And the 2 bits of Sodium

EDIT 2: Ok, gonna try again....uh...Sulfate and Hydroxide ions are the Spectator Ions? And the total ionic equation would be Zn(OH)2 + NaSO4? Am I correct in my guesses?
Title: Re: Zinc sulfare + Sodium hydroxide questions
Post by: Borek on November 21, 2013, 04:41:38 PM
Let's start from the beginning. Explain how you got from

2NaOH + ZnSO4  :rarrow: Na2SO4 + Zn(OH)2

to

Zn2+ + 2OH- :rarrow: Zn(OH)2
Title: Re: Zinc sulfare + Sodium hydroxide questions
Post by: dudebuddyguy on November 21, 2013, 04:44:56 PM
Quote from: Borek on November 21, 2013, 04:41:38 PM
Let's start from the beginning. Explain how you got from

2NaOH + ZnSO4  :rarrow: Na2SO4 + Zn(OH)2

to

Zn2+ + 2OH- :rarrow: Zn(OH)2

Zn2+(aq) + 2 OH-(aq) -> Zn(OH)2(s)

This? Googled it. I told you, grade 10 science. Kicking myself for not getting grade 11/12 back in high school.
Title: Re: Zinc sulfare + Sodium hydroxide questions
Post by: Borek on November 21, 2013, 05:02:56 PM
Dissociation in water solutions:

NaOH :rarrow: Na+ + OH-

ZnSO4 :rarrow: Zn2+ + SO42-

Bases generally dissociate into metal cation and OH-, acids into H+ and an anion (for example Cl-), salts into cation and anion.
Title: Re: Zinc sulfare + Sodium hydroxide questions
Post by: dudebuddyguy on November 21, 2013, 05:05:46 PM
Quote from: Borek on November 21, 2013, 05:02:56 PM
Dissociation in water solutions:

NaOH :rarrow: Na+ + OH-

ZnSO4 :rarrow: Zn2+ + SO42-

Bases generally dissociate into metal cation and OH-, acids into H+ and an anion (for example Cl-), salts into cation and anion.

Ah, that makes sense. So Na+ + OH- + Zn2+ + SO42- is the total ionic equation, I assume?
Title: Re: Zinc sulfare + Sodium hydroxide questions
Post by: Borek on November 21, 2013, 05:11:14 PM
Almost. You need to write correct amounts of all ions (so 2NaOH becomes 2Na+ + 2OH-).
Title: Re: Zinc sulfare + Sodium hydroxide questions
Post by: dudebuddyguy on November 21, 2013, 05:15:34 PM
Quote from: Borek on November 21, 2013, 05:11:14 PM
Almost. You need to write correct amounts of all ions (so 2NaOH becomes 2Na+ + 2OH-).

Beauty! I'm learning :D