Question

HOW AM I SUPPOSED TO CALCULATE THIS??
Formic acid, HCOOH, which causes the sting felt from some insect bites, is 1.48% dissociated in 0.50 M HCOOH (aq) at a certain temperature.
a. Calculate [H+].
b. Calculate Ka for HCOOH at this temperature.
c. Determine pH of 0.50 M HCOOH.

Answer 1

\[HCOOH \leftarrow \rightarrow H^+~~HCOO^{-}\]

\[K_{a} = \frac{ [HCOO^-][H^+] }{ [HCOOH] }\]

Answer 2

do i have to substitute their molar mass ? @Photon336

Answer 3

which part are you working on right now?

Answer 4

I have not started @aaronq photon said he was going to come back in a while . I have no clue how to do this

Answer 5

a) is pretty simple, you know that the acid dissociates like: \(\sf HA\rightarrow H^++A^-\), so just find 1.48% of the initial concentration of the acid (0.5 M) and thats equal to \([H^+]\).

for b) you would use the info from a). the concentrations are:

\(\sf [H^+]=[A^-]\) and \(\sf [HA]_{eq}=[HA]_0-[H^+]\)

for c) use \(\sf pH=-log[H^+]\)

Answer 6

If you need to visualize b) further, you can make an I.C.E table (initial, change and equilibrium concentrations).

Answer 7

I'm not that bright of a student. i need a simpler explanation. i appreciate your effort :) i just dont know how to do this at all @aaronq

Answer 8

oh dont say that, you probably just havent really applied yourself :P

so for a) it's a simple concept. if i asked you what is 5% of $1.00, what would you say?

Answer 9

.05?

Answer 10

right! how did you figure that out?

Answer 11

i changed 5% to 5/100 and multiplied it into 1

Answer 12

exactly. so what is 1.48% of 0.5 M?

Answer 13

.0074

Answer 14

yup! so \(\sf [H^+]=0.0074~M\)

Answer 15

yay :) what is the formula for b?

Answer 16

this goes back to dissociation equation: \(\sf HCOOH\rightarrow H^++HCOO^-\)

Answer 17

what photon wrote it the equilibrium expression for the dissociation. something you would need to know how to write if you were going to the question alone.

We need to make an I.C.E. table, we start off with 0.5 M of the acid and none of the products.

\( \sf [HCOOH] ~~~~~~ [H^+]~~~~[HCOO^-] \)
I 0.5 M 0 0
C
E

then the change is, 0.0074, which we will represent as x for now.

\( \sf [HCOOH] ~~~~~~ [H^+]~~~~[HCOO^-] \)
I 0.5 M 0 0
C -x +x +x
E

finally we equilibrium concentrations are just putting initial and change values together.

\( \sf [HCOOH] ~~~~~~ [H^+]~~~~[HCOO^-] \)
I 0.5 M 0 0
C -x +x +x
E 0.5 -x x x

We then use these in the equilibrium expression, try plugging them in.

Answer 18

Whoa

Answer 19

it's not complicated, i just wrote one step at a time to help you see what you do at each step.

Answer 20

which one do i use though

Answer 21

the equilibrium (E) values

Answer 22

Um ok

Answer 23

I'll give you a hand. We have the equilibrium expression, so we plug in the values from E.

\(\sf K_{a} = \dfrac{ [HCOO^-][H^+] }{ [HCOOH] }=\dfrac{x*x}{0.5-x}=\dfrac{x^2}{0.5-x}\)

but remember we found out what x was in part a)? \(\sf [H^+]=x=0.0074\)

\(\sf K_a=\dfrac{x^2}{0.5-x}=\dfrac{(0.0074)^2}{0.5-0.0074}\)

Answer 24

I got such a huge number
@aaronq

Answer 25

it should be a very small number, remember this is the dissociation constant of a weak acid that only dissociates 1.48%.

Answer 26

but i keep getting 0.00011116 what i meant by huge is it has many number not a Big number

Answer 27

ohh that's actually the right answer!

um huge and big are synonyms, btw :P haha

try writing that number in scientific notation though, it's usually how it's expressed and you should be able to write it as so.

Answer 28

Lol i feel so dumb atm

Answer 29

so is it 1.1116 × 10-4 @aaronq

Answer 30

\(\huge \checkmark\)

Answer 31

yes! so that was b right

Answer 32

yes, that was b)

Answer 33

for c) use \(\sf pH=-log[H^+]\)

you already know \(\sf [H^+]\) from a)

Answer 34

ohhh okay thank you :) :)

Answer 35

no problm (: