Question

The pH of a juice drink is 2.6 Find the concentration of hydrogen ions in the drink.
a.2.6
b.2.5*10^-3
c.10^-2.6
d.2.5*10^-3

Answer 1

do you have the formula?

Answer 2

No I don't . :/

Answer 3

pH = -log_10[H+]

Answer 4

I should really have that formula memorized oh well

Answer 5

that is the formula, do you know how to solve it now?

Answer 6

http://www.ausetute.com.au/phscale.html

Answer 7

I can show you how to solve it if the answer is no

Answer 8

Also note
pH = -log[H+]
as it is log base 10

Answer 9

Still confused and the answer is no .

Answer 10

alright note that

log(x) = y
is the same as
x = 10^(y)

Answer 11

Okay

Answer 12

Also note that

10^(log(x)) = x
7^(log_7(x))
etc..

pH = -log_10[H+]

So we know the pH is equal to
pH = 2.6
thus

2.6 = -log_10[H+]
multiply both sides by -1
-1 * 2.6 = -log_10[H+] * -1
-2.6 = log_10[H+]
Now make both sides of the equation exponents to the base 10 to eliminate the logarithm

10^(-2.6) = 10^(log_10[H+])
10^(-2.6) = [+H]

REMEMBER SQUARE BRACKETS MEAN CONCENTRATION (probably Molarity)
so we know the concentration of +H is 10^(-2.6) M

Answer 13

I hope this helps you, logarithms are pretty important to understand I think at least for solving some basic chemistry formulas

Answer 14

Chemistry formulas are important to understand but I got this question on Algebra 2 HW

Answer 15

Weird to get a question with a specific formula like this, although I'm sure it makes sense if you analyze the question.

Logarithmic rules
1) log_b(mn) = log_b(m) + logb(n)

2) log_b(m/n) = log_b(m) – logb(n)

3) log_b(m^(n)) = n · log_b(m)

4) log_3(3) = 1

5) log_e(x) = ln(x)

6) log(x) = y is the same as x =10^(y)

7) log_10(x) = log(x)

8) log_e(x) = ln(x)

9) e^(ln(x)) = x

Answer 16

here are the rules for logarithms hope they help you

Answer 17

do you understand how to solve it now?

Answer 18

Note that
10^(-2.6) = 0.0025M which is a pretty high concentration of H+ ions which is expected of a solution that has a pH that low. So this answer is reasonable

Answer 19

So 4 log x - 6 log (x+2) is an example of rule #2 ?

Answer 20

Ya I got that on my TI .

Answer 21

log_b(m/n) = log_b(m) – logb(n)

it is an example if the original logarithm was

log(x^(4)/(x+2)^(6))

Answer 22

as you can simplify it down to what you have written there

Answer 23

But I have to simplify it to a single logarithm .

Answer 24

so just follow my rules start with the logarithms individually

Answer 25

or the rules they aren't mine :)

Answer 26

sorry I made a mistake on rule 2 might have made it confusing

2) log_b(m/n) = log_b(m) – log_b(n)

Answer 27

4 log x - 6 log (x+2)
for the first one
see that

log(x^(4)) = 4log(x)

and

log((x+2)^(6)) = 6log(x+2)

Answer 28

Notice the negative in front of the logarithm indicates that the exponent is negative
- 6 log (x+2)
so

- 6 log (x+2) = log((x+2)^(-6))

This is significant because

x^(-1) = 1/x
x^(-56) = 1/x^(56)
1/x^(-1) = x/1
etc


so we know right away when we combine these logarithms that it is going to be in the denominator

Answer 29

The last comment is just for interest you can blindly apply the rules but understanding why things are the way they are is always nice

Answer 30

I put it into my calculatorr and got -2.5

Answer 31

what?

Answer 32

I put that problem into my calculator and came out with -2.5