Destroyer GChem #17

[hopeful dental]

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Hi guys,

I know I should know this, but I don't and need some help....I understand #17 until I get up to the calculation 8x10(-7) (-7 is the exponent)=x(2) (2 is exponent)=80x10(-8)=x(2)=9x10(-4)=x=OH- ????????

I hope this makes sense....please help....so confused!!


Really appreciate it!

 

[heatwheat]

Well you know pOH = -log [OH]

So you need to solve for [OH] in order to solve for the pOH, and then you can solve for the pH.

Once you get to : (8x 10^-7) = X^2, you solve for X by taking the square root of both sides. X = [OH]

Since you have [OH], plug it into pOH = -log[OH], then use the equation
14 = pH + pOH, and solve for pH.

 

[heatwheat]

Well you know pOH = -log [OH]

So you need to solve for [OH] in order to solve for the pOH, and then you can solve for the pH.

Once you get to : (8x 10^-7) = X^2, you solve for X by taking the square root of both sides. X = [OH]

Since you have [OH], plug it into pOH = -log[OH], then use the equation
14 = pH + pOH, and solve for pH.

 

[hopeful dental]

Thanks Heatwheat.....I knew it seemed simple, and when I didn't hear back from anyone, I kept looking at it, then putting it down and looking back at it...then it finally hit me...gosh, I feel stupid!!!

Anyway, thanks a bunch...I'm sure I will have more questions....hopefully they are not as stupid as this one!

 

[heatwheat]

No problem. BTW, I have no idea why my post cam up twice, with like a couple hours in between... Oh well. Glad you figured it out.