Neutralizing HCl solution with Sodium Bicarbonate..

[JARY]

So basically this is the situation. Gonna be acid treating some starch with some dilluted hydrochloric acid solution, then before we pitch it we gotta neutralize it with baking soda, sodium bicarbonate. I think baking soda's pH ~8? Anyways, just need to figure out how many normal NaHCO3 vs how many Normal of HCl to add.

 

Solution is gonna be:

 

25 mL H20 (heated)

30 mL HCl (added)

= 55 mL total

 

If I could get some help in how to approach this that would be great, I'm still a little ambiguous. If I'm missing details sry, just maybe give me an overall idea of how to do it lol

 

I'm supposed to use Normality too, of HCl in comparison with normality of the NaHCO3, thanks >.>

[Melvin]

It depends on the concentration of the HCl that you added (its not pure HCl, which is a gas).

[hermanntrude]

what you need to do is this:

 

1) find out how many moles of HCl you have

2) use the chemical equation for the neutralisation to determine how many moles of bicarb you need

3) calculate what mass of bicarb that equates to

 

step 1 I cannot help you with unless you know the concentration of your HCl. If you can find that out, I (and many others on the site can help you with this calculation. it's simple).

 

step 2 I can help you with. the equation is this:

 

[ce] HCl + NaHCO3 --> NaCl + H2O + CO2[/ce]

 

What this equation tells us is that one mole of HCl reacts with one mole of bicarb. In other words, however many moles of HCl you have, you need the same quantity of bicarb.

 

step 3 will need another calculation. Whatever number of moles of bicarb you require, multiply by the molar mass of bicarb (84 g/mol). This is your answer in grams.

[JARY]

Thanks!

 

Apparently it is 6 N HCl solution. If that helps :\

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I've tried a few calculations out... naw, that can't be right what I'm getting;

 

Normality = Molarity * n (valence), since HCl only gives up one ion, molarity should also be 6 M, right.

 

(0.3 L HCl)*(6 mol/L) = 1.8 mol HCl = 1.8 mol NaHCO3 (84g NaHCO3/mol) = ??? That's way too high, I think I'm missing something, like some dillution err... lol. Just doing it the way you described.

 

I also know N = (weight solute/equiv W) / (Volume). To me that pretty much translates like molarity, just factoring in the ions from H+ or OH-

Edited September 22, 2009 by JARY
Consecutive posts merged.

[hermanntrude]

your calculations are perfect in every respect except for your conversion of 30ml into L. 30mL is 0.03L, not 0.3.

 

you are correct in your understanding of "6N"

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i make it a bit more than 15g.

 

be careful when you do this. the fizzing will mean a lot of HCl fumes are released, and you have a fairly concentrated solution. Do this ina very well ventilated area, protective clothing, safety glasses and stand well back.