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Mathematics 13 Online
OpenStudy (cloverracer):

Need help with algebra 2 question! See attachment.

OpenStudy (cloverracer):

OpenStudy (mathmale):

Please look at the given formula. Copy it down. Note that [H+] is given and is 4.3*10^(-14).

OpenStudy (mathmale):

Just substitute this VALUE for [H+] into the formula, calculate the log, and then put a negative sign in front of it.

OpenStudy (cloverracer):

so pH = -log4.3*10^(-14)

OpenStudy (cloverracer):

then pH=-log4.3-log10^(-14) ?

OpenStudy (mathmale):

Enclose the argument (4.3 ....) in parentheses. Otherwise you're doing fine. You'll need to evaluate that log. Then, you'll need to negate the value of the log (put a - sign in front of it).

OpenStudy (cloverracer):

I'm stuck on evaluating the log -log4.3

OpenStudy (cloverracer):

I know -log10^(-14) is just (-14)

OpenStudy (mathmale):

You have too many - signs in your expression. Only one of them belongs there. CAn you fix this? As before, evaluate log (..... ) before writing the neg. sign in front of it.

OpenStudy (cloverracer):

(-14)?

OpenStudy (mathmale):

[H+] is given and is 4.3*10^(-14). 4.3 * 10^(-14) is a product. What is the product rule for logs? True, log 10^(-14) is -14.

OpenStudy (cloverracer):

logbAC = logbA + logbC

OpenStudy (mathmale):

Yes, that's correct. So, whats the log of the product 4.3 * 10^(-14) ?

OpenStudy (cloverracer):

ehh, i'm stumped

OpenStudy (mathmale):

What's the log of 4.3? What's the log of 10^(-14)? Whats the log of the product of those two quantities?

OpenStudy (cloverracer):

oh wait is it -0.63347?

OpenStudy (cloverracer):

for -log4.3

OpenStudy (mathmale):

I was hoping you'd evaluate the 2 logs and then add the results together before brining in that negative sign. You have already told me, correctly, that log 10^(-14) is -14. What is the log of 4.3?

OpenStudy (cloverracer):

the log of 4.3 is 0.63347 right?

OpenStudy (mathmale):

Yes, that looks good. Now, add to gether that 0.63347 and that -14. Can you explain why this is necessary?

OpenStudy (mathmale):

Note that one is positive and the other negative.

OpenStudy (cloverracer):

would you solve it like log0.63347 * -14

OpenStudy (mathmale):

No, you no longer need that "log" operator, because you've already found the logs of 4.3 and 10^(-14).

OpenStudy (cloverracer):

ohh okay so it's just 0.63347 * -10?

OpenStudy (mathmale):

This isn't multiplication, so throw out that *. Combine (do not multiply) 0.63347 and -14.

OpenStudy (mathmale):

* means multiplication + denotes addition

OpenStudy (cloverracer):

I got -9.366..

OpenStudy (cloverracer):

oh wait careless mistake, now my answer is -13.366..

OpenStudy (mathmale):

much better. Now negate that result (multiply it by -1). \

OpenStudy (cloverracer):

it is it's reciprocal, 13.3?

OpenStudy (mathmale):

No. You have -13.366, which is a negative number, and need only multiply that negative number by -1. what is the result of that? It's true that an EXPONENT of -1 indicates reciprocal in some cases, but not in this one.

OpenStudy (cloverracer):

the result of -13.366 * -1 is 13.366?

OpenStudy (mathmale):

right, and that's to 3 decimal places. How many decimal places were called for?

OpenStudy (cloverracer):

it rounds to 13.4 right?

OpenStudy (mathmale):

Right. That's rounded off to the nearest 10th. Your answer? "The approximate pH is ... ?"

OpenStudy (cloverracer):

The appoximate pH is 13.4

OpenStudy (mathmale):

Right. Very good. Any questins about this procedure?

OpenStudy (cloverracer):

Not anymore, thank you so much for your help!

OpenStudy (mathmale):

You're welcome. Best of luck to you, and M. C. Bye!

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