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Ch 6.6 Weak acids

Ch 6.6 Weak acids

Postby JWC01 » Tue Jul 31, 2012 8:25 am

For the calculation of -log 1.32 x 10-4 =3.88.

How do you do this without a calculator? Am i missing something here?
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Postby goldstanda3269 » Tue Jul 31, 2012 11:53 pm

You can't calculate it precisely without a calculator but you can calculate it "good enough" in order to choose the right answer in a multiple choice exam. This is explained near the end of section CHM 6.6.1.
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Postby JWC01 » Wed Aug 01, 2012 6:53 am

Lovely, thanks for your help.
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Re: Ch 6.6 Weak acids

Postby jk_vandyk8141 » Tue Jul 10, 2018 8:55 pm

I am having troubling figuring out the math steps involved in calculating x in the ICE question on page 80. I am unsure how you got from x^2/10^-2 = 1.75 x 10^-5 to x =4.18 x 10^-4.

I think i am missing a crucial step at the square root step.

Thanks
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Re: Ch 6.6 Weak acids

Postby goldstanda3269 » Wed Jul 11, 2018 2:16 am

jk_vandyk8141 wrote:I am having troubling figuring out the math steps involved in calculating x in the ICE question on page 80. I am unsure how you got from x^2/10^-2 = 1.75 x 10^-5 to x =4.18 x 10^-4.

I think i am missing a crucial step at the square root step.


x^2/10^-2 = 1.75 x 10^-5

multiply both sides by 10^-2 so that it cancels with the denominator on the left side

(10^-2) x^2/10^-2 = 1.75 x 10^-5 (10^-2)

x^2 = 1.75 x 10^-7

Without a calculator, it is not convenient to take the square root of an uneven exponent (i.e. 10^-7) so we should adjust the scientific notation so that there is an even exponent (GAMSAT Math chapter 1, GM 1.5).

x^2 = 1.75 x 10^-7 = 17.5 x 10^-8

Keep in mind that 4 times 4 is 16 so the square root of 16 is 4. We can estimate, especially in a multiple choice exam, the square root of 17.5 as approx. 4.2.

x = approx. 4.2 x 10^-4

x = 4.18 x 10^-4
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