Solving for x in Problem 12.59 a

Acidity
Basicity
The Conjugate Seesaw

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Katie Clark 3B
Posts: 26
Joined: Fri Sep 25, 2015 3:00 am

Solving for x in Problem 12.59 a

Postby Katie Clark 3B » Fri Dec 04, 2015 12:30 pm

When calculating the pH and pOH of 0.057 M NH3, I am confused about an approximation that the solution manual made when solving for x in the Kb equation. The chemical equation was NH3(aq)+H20(l)=NH4(aq)++OH-(aq). I did an ice box and had that Kb=x2/(0.057-x). The answer in the solution manual had said that this could be approximated to x2/0.057. How do you know when you can make this approximation in the Kb or Ka equations? Can someone explain this to me?

Rex Lee 1C
Posts: 48
Joined: Fri Sep 25, 2015 3:00 am

Re: Solving for x in Problem 12.59 a

Postby Rex Lee 1C » Fri Dec 04, 2015 12:37 pm

The approximation is made because the Kb for Ammonia (1.8e-5) is so small. Because of this, the x value that you would get when you solved the quadratic would be very small in relation to the starting concentration. Since the x value is so small, you can reduce (0.057-x) to 0.057 in the denominator because the concentration change won't be significant.


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