Solving Square Roots for Exponential Forms in pH, etc

[DATGal]

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I am struggling through pH, pKa material because of the math. Namely how to “guesstimate” the square roots of the exponential forms. For example (see below), I got up to -log (√1x10-5) 👍 for the problem below but from there, I don’t know how to get the correct answer in pH ! 😕 Any additional examples in addition to helping this would be highly appreciated. Thank you!!!

The Ka of an acetic acid is approx. 2x10-5. What is the pH of a 0.5M solution of acetic acid?😕

 

[keliao]

let me try
CH3COOH + H2O -------> CH4COOH + OH i am not sure if i have CH4COOH rite...some1 correct me.

Ka= [CH4COOH] [OH] / [CH3COOH]

2E-5 = X^2/ 0.5M solve for X
X = 1E-5 pH is less than 5 (alwyas look at the negative number) so guessing
pH is 4.8

 

[waystinthyme]

let me try
CH3COOH + H2O -------> CH4COOH + OH i am not sure if i have CH4COOH rite...some1 correct me.

Ka= [CH4COOH] [OH] / [CH3COOH]

2E-5 = X^2/ 0.5M solve for X
X = 1E-5 pH is less than 5 (alwyas look at the negative number) so guessing
pH is 4.8

didn't you skip a step here? i see where you multiplied both sides by 0.5, but what happend to your x^2? i am not good with pH calculations either, so any help would be greatly appreciated...

-waystinthyme

 

[DATGal]

Okay, I think I understand it now. By the way, the answer is: pH= 2.5

Q. The Ka of an acetic acid is approx. 2x10-5. What is the pH of a 0.5M solution of acetic acid?

Ka = [H+][A-]/[[HA]
2x10^ -5 = x^2 /0.5
x^2 =10x10^-6 or 1.0x10^5
x= √1x10-5
To convert to pH, take the – log of [H+]
-log (√1x10-5)
Recall that log√ x = (1/2) log x and that –log subscript (10-x) = x
In other words, divide the exponent by 2 (and if it is a cube root, divide by 3) and then you get 2.5 as the pH.

Hopefully, this is helpful to others struggling with it as well.

 

[waystinthyme]

good work.

i was a little confused by the second step, but now i think i get it. to make sure though, you just used a conversion there, right?

0.5 = 5 x 10^-1, so:

2 x 10^-5 = x^2 / 0.5
(2 x 10^-5) (5 x 10^-1) = (x^2 / 0.5) (0.5)
10 x 10^-6 = x^2

and so, on...is that the right rationale for that step?

-waystinthyme

 

[chessxwizard]

Or when you're at the x^2 = 10 x 10-6 step, you can realize that 3^2=9 and 4^2=16, so that the root would be really close to 3.

x = 3 x 10^-3.

-log (3 x 10^-3) is between 3 and 2, most likely around 2.6 since the first number is a little larger than 1.

 

[waystinthyme]

Or when you're at the x^2 = 10 x 10-6 step, you can realize that 3^2=9 and 4^2=16, so that the root would be really close to 3.

x = 3 x 10^-3.

-log (3 x 10^-3) is between 3 and 2, most likely around 2.6 since the first number is a little larger than 1.

this is probably a dumb question, but how do you know it's between 3 and 2 at this point? can you break this step down a little more?

-waystinthyme

 

[chessxwizard]

Let's take, for example, a x 10^c, where a is between 1 and 10

The square root of that is between -c and (-c - 1), and a basically tells u how close you are to each extreme. For instance, when a is 1, then you have -c; when a is 10, then you're at (-c -1).

ie. 1 x 10^-3. -log of that is 3.
10 x 10^-3. -log of that is 2.
5 x 10^3. Somewhere in between. Since it's on a log scale, you can't be completely sure, but you can estimate.

 

[sunshine222]

I found a pretty good (and accurate) technique online somewhere and it's what I used, although you do have to memorize a few values.
Basically just memorize the values of log(x) for 0<x<10 - they're not too hard to remember:

log(1) = 0
log(2) = .3
log(3) = .48
log(4) = .6
log(5) = .7
log(6) = .78
log(7) = .85
log(8) = .9
log(9) = .95

Then when you have to take a log, say -log (.0004), do this:

-log(.0004)

= -log(4 x 10^-4)

= -log(4) + -log(10^-4)

= -.6 + 4

= 3.4

the actual value is 3.398, so you're getting a pretty accurate result.

 

[supraman]

Whenever its in this form:

N x 10&#713;ª

the -log [N x 10&#713;ª] is simplified into:

N- (log a)

just memorize three [log a]

log 2 ~ 0.3
log 5 ~ 0.7
log 8 ~ 0.9
and base other logs in between those (although the values will not be exactly in between)

So then for example

5 x 10&#713;³

Finding its log is:

3 - log 5
3 - [0.7] = 2.3

Hope that helps! Sorry if everyone already stated this lol

 

[chessxwizard]

Building on sunshine's post...

You can use the addition/multiplication rules of logs to make the list of numbers that you have to memorize even smaller...

ie. log 6 = log (2*3) = log 2 + log 3.
log 8 = log (2*2*2) = log 2 + log 2 + log 2
etc...

 

[supraman]

Also if you have to divide for example something wack like this:

[ 1 x 10&#713;¹³] / [5 x 10&#713;³]

Break it up into two fractions:

[1/5] and [10&#713;¹³ - 10&#713;³]

[1/5] = 0.2 or 2.0 x 10&#713;¹
[10&#713;¹³ - 10&#713;³] = 10&#713;¹³&#713;&#713;³ = 10&#713;¹º

To combine, they have to be multiplied because they were originally seperated when they were multiplied together (remember multiplying you ADD exponents, dividing you subtract).
2.0 x 10&#713;¹ x 10&#713;¹º = 2.0 x 10&#713;¹¹

Of course the matter gets harder when you have to divide say 1/6.2, but you can always round to 1/6 and such.

 

[waystinthyme]

those were all really helpful tips...thanks!

-waystinthyme