Solving for x in ICE table

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JulieAljamal1E
Posts: 71
Joined: Fri Sep 28, 2018 12:24 am

Solving for x in ICE table

Postby JulieAljamal1E » Sat Jan 12, 2019 4:36 pm

On question 5I.17 in the 7th ed., there is a step in the solutions manual that goes from 2x=(3.16x10^-3)(0.114-x) followed by the step 2.00316x=3.60x10^-4. Can someone explain the work process in between those steps or how to get from that first step to the next?

Kavvya Gupta 1H
Posts: 35
Joined: Fri Sep 28, 2018 12:21 am

Re: Solving for x in ICE table

Postby Kavvya Gupta 1H » Sat Jan 12, 2019 4:41 pm

you multiply (3.16*10^-3) with (0.114-x) getting 2x= (3.6*10^-4)-(3.16*10^-3)x. Then you get the x on one side getting 2.00316=3.6*10^-4

105114680
Posts: 60
Joined: Fri Sep 28, 2018 12:23 am

Re: Solving for x in ICE table

Postby 105114680 » Sat Jan 12, 2019 4:43 pm

Multiply the right hand of the equation out. Doing so gives you 2x=3.60x10^-4 - 3.16x10^-3x (multiply 3.16x10^-3 in the first parenthesis with both terms in the second parenthesis). To solve for x add the 3.16x10^-3x from the right hand side of the equation to the 2x on the left. This gives you 2.00316x=3.60x10^-4. To isolate x from here, divide both sides by 2.00316. Hope this helped!


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