Question

how to solve for
p[oH]
p[H]
on a calculator
(i understand how it works but the numbers do not come out correctly)

Answer 1

it depence on the calc...
try:
log
value
=

Answer 2

hmm i just have to plug in -
then log
it prints out -log (

Answer 3

its a TI 30x :/

Answer 4

lets try something..
type:
log
100
=

Answer 5

what is it say?

Answer 6

2

Answer 7

log
0.001
=
?

Answer 8

-3

Answer 9

ok

Answer 10

-
log
0.001
=
?

Answer 11

3

Answer 12

working good
so:
-
log
(value of [H] )
=
.....

Answer 13

thats the part i do not understand

OH = 3.0 * 10 ^-9



example H = 3.3 * 10^-6

how did the book get OH ?

Answer 14

vice versa also :/

Answer 15

pOH + pH = pKw

Answer 16

[H+] *[OH-] = Kw

Answer 17

Kw=?

Answer 18

lets take it from the top...
what you know
what to find?

Answer 19

that is the yheory
mumbers...???

Answer 20

i only got

1.0 * 10^14 = 3.3 * 10^-6 [OH]

Answer 21

but how do i punch it in on a calc?

Answer 22

1.0*10^-14 seems like Kw
so
Kw = [ H+] [ OH-] =>
[ H+]= Kw / [ OH-] =>
[ H+]= 1.0 * 10^-14 / 3.3 * 10^-6 =>
[ H+]= 0.3 * 10^-8 =>
[ H+]= 3 * 10^-9

so
pH=-log[ H+] =>
pH=-log 3 * 10^-9
pH=8.5

Answer 23

punch:
1
exp
-
14
/
3.3
exp
-
6
=
log
=
+-

Answer 24

is it working???