Writing equations

how would I do this @doug.walker?

Hey Sarah

NaHCO3 is an ionic compound which means it will dissociate into it’s ions in solution so the equation will be

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The follow up from this would be the “HCO3–” ion which is amphiprotic (meaning it can act as an acid or base).

The definition for an acid is a proton donor so if it is acting as an acid it will donate it’s proton to water, which then increases the concentration of hydronium ions in solution (and therefore making the solution acidic - ref to the first equation)
Bases are proton acceptors, so if it is acting as a base, (ref to the second equation), it will accept a proton from water, resulting in the production of the hydroxide ion. This would then increase the pH, making the solution basic.

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Hope that helps.
*Apologies for the typed up chemical ions, I can’t get the subscription function to work properly on here.

thank you so much this really helped!!

Thanks for the response.
I would also emphasis that you notice that in the first equation showing the initial dissolving process with water that it is important to use a single headed arrow as you have correctly done. This means that the substance is 100% dissociated into ions (as all soluble ionic salts and all strong acids and bases will be).
In contrast, the next reaction which involves the acid-base reaction of the bicarbonate ion with water uses double headed (equilibrium) arrows as any acidic or basic ion will only ever be a weak acid or base and therefore will only be partially dissociated in water.
If the ion involved was chloride there would be no 2nd reaction as Cl- is the conjugate of the very strong hydrochloric acid so is far too weak a base to have an impact on the pH. If the ion had been CH3COO- which is the conjugate of the weak ethanoic acid then the solution would be a weak base because of the 2nd reaction CH3COO- + H2O → CH3COOH + OH-
In the bicarbonate example above, the answer is complicated because it is amphiprotic which means sometimes it acts as a base and sometimes as an acid , depending on what it is reacting with. In this case it turns out that the bicarbonate ion is a stronger base than water is, so a slightly basic solution forms. But an understanding of that choice is not expected until Yr 13. :slight_smile: