Calculating Ka

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Quinton Sprague 1A
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Calculating Ka

Postby Quinton Sprague 1A » Wed Dec 09, 2020 10:53 am

In today's lecture, Dr. Lavelle discussed how H2CO3 can actually donate both of its H+ in a reaction with water. In calculating the Ka, he showed how you multiply the products in the numerator and divide this by the solvent. Wondering if actually calculating ourselves, what numbers we put into this equation? Do we use the molarity of products and solvents, or the moles of each? Assuming we will not be calculating this on the final just curious...

Aliya 1H
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Re: Calculating Ka

Postby Aliya 1H » Wed Dec 09, 2020 10:59 am

Ka calculations use concentrations aka molarity, which is indicated by the square brackets [ ] whenever you see like the Ka written like . Also, I believe it's only aqueous compounds from your chemical reaction that are included in Ka.

Rob Tsai 2F
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Re: Calculating Ka

Postby Rob Tsai 2F » Wed Dec 09, 2020 3:04 pm

In this example, you would divide the product of the concentrations of the products over the concentration of the solvent:

AH <—> A- (aq) + H+ (aq) leads to Ka =( [A-][H+] )/[AH]

Concentrations, if they aren't given, would be calculated through our molarity formula M = n/V, where n=moles and v= volume in liters.

Michael Sun Dis 3G
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Re: Calculating Ka

Postby Michael Sun Dis 3G » Thu Dec 10, 2020 4:53 pm

Since Ka requires the concentrations of relevant reactants and products, you would need to use molarity.

Joshua Eidam 2A
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Re: Calculating Ka

Postby Joshua Eidam 2A » Fri Dec 11, 2020 12:02 am

So what is the exact equation for Ka? Also if only aqueous samples are used in calculating the Ka, then are any samples in different forms (solid, liquid, gas) just disregarded?

Jordi M 2I
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Re: Calculating Ka

Postby Jordi M 2I » Fri Dec 11, 2020 12:06 am

Rob Tsai 2F wrote:In this example, you would divide the product of the concentrations of the products over the concentration of the solvent:

AH <—> A- (aq) + H+ (aq) leads to Ka =( [A-][H+] )/[AH]

Concentrations, if they aren't given, would be calculated through our molarity formula M = n/V, where n=moles and v= volume in liters.


Wouldn't the solvent be water? I think you mean concentration of solute since we never include water in the Ka calculation.

AHUNT_1A
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Re: Calculating Ka

Postby AHUNT_1A » Fri Dec 11, 2020 8:06 am

Using this as reference !

Gigi Elizarraras 2C
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Re: Calculating Ka

Postby Gigi Elizarraras 2C » Fri Dec 11, 2020 9:16 am

You would use the molarities of the products and the reactants so to calculate KA you need to know the concentrations:)

Lorraine Jiang 2C
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Re: Calculating Ka

Postby Lorraine Jiang 2C » Sat Dec 12, 2020 12:20 am

Hi! The formula for Ka will be [A-][H3O+]/[HA] so that is the concentration of the anion multiplies the concentration of the H+ ion divided by the concentration of the acid.

Hope it helps!

Rohit Srinivas 2D
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Re: Calculating Ka

Postby Rohit Srinivas 2D » Wed Dec 16, 2020 9:08 pm

We add in molarity (concentration). We haven't done kinetics yet but whenever we see the [] that indicates concentration.


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