Question

What is the pH of a solution that is prepared by mixing 100 mL of 0.50 M sodium acetate and 115 mL of 0.100 M acetic acid?

@photon336

Answer 1

@agent0smith (one of the best for science)

Answer 2

\(NaOH+CH_3COOH \rightarrow \text{Solution}\)

pH of a solution = \(-log[H^+]\)

Answer 3

\(CH_3COOH \rightarrow CH_3COOH^- + H^+\)
Initial concentration= 0.100M 0 0
Final concentration= 0.100(1-\(\alpha\) ) \(0.100\alpha\) \(0.100 \alpha\)

Answer 4

\[K_a= \frac{(0.110 \alpha)(0.110 \alpha)}{0.110(1- \alpha)}\]where \(k_a\) is the dissociation constant for acetic acid & is equal to \(1.75 \times 10^{-5}\).

Answer 5

from here u may find \([H^+]=-0.110 \alpha\)

& then u may find pH of the solution

Answer 6

@jiteshmeghwal9 what does alpha refer to?

Answer 7

n acid = 115 x 0.100 = 11.5 mmol
n base = 100 x 0.50 =50 mmol

H+ = Ka *acid/ base
H+ = 1.75 × 10 −5 * 11.5/50
H+ = 4 x 10^-6

pH = - log [H+]
pH = - log 4 x 10^-6
pH = 5.40

Answer 8

@Summersnow8 What do u think?

Answer 9

where did you get the ka?

Answer 10

I'm searching...

Answer 11

We don't have [H+]
So I think we can't find Ka

Answer 12

Should we find it using formula?

Answer 13

I am not sure. where did you get 1.75 × 10 −5

Answer 14

searching on google

Answer 15

It said acetic has 1.75 x 10^-5

Answer 16

hmmmmm, well that won't work. I need to be able to solve it without the computer

Answer 17

What formula we can use to find Ka?

Answer 18

Ka = ( [H+] [A¯] ) / [HA]

Answer 19

What is A- and HA ?

Answer 20

Sorry I'm asking since I really don't know what is it. In my country, maybe it written in other word.

Answer 21

A would be the base, while HA would be an acid

Answer 22

That's what we're looking for, the [H+]
I will ask someone about this question, maybe he can help.

Answer 23

okay