PH for this Weak Acid-Strong Base titration.

[Ace.]

$\:$

Homework Statement

What is the pH when 10 ml of 0.1 M $NaOH$ is added to 25 ml of 0.1 M $H_{}C_{2}H_{3}O_{2}$?

Homework Equations

pH = -log[$H_{3}O$]

The Attempt at a Solution

The main thing confusing me about this type of problem is that my teacher taught me a way which is different from the methods I see online. And I get a different answer.

This is how I was taught to do it:

moles of $H_{}C_{2}H_{3}O_{2}$ = c * v
= 0.1 M * 0.025L
= 0.0025 mol

moles of $NaOH$ = c * v
= 0.1M * 0.01L
= 0.001 mol

Ethanoic acid is in excess, so
moles $H_{}C_{2}H_{3}O_{2}$ = 0.0025 mol - 0.001 mol
= 0.0015 mol

[$H_{}C_{2}H_{3}O_{2}$] = $\frac{n}{v}$ = $\frac{0.0015}{0.035}$ = $0.043$

$H_{}C_{2}H_{3}O_{2} + H_{2}O \leftrightharpoons H_{3}O + C_{2}H_{3}O_{2}^{-}$
I $\:\:$ 0.043 $\:\:\:\:\:\:$ - $\:\:\:\:\:\:\:\:\:\:\:\:$ 0 $\:\:\:\:\:\:\:\:\:\:\:\:$ 0
C $\:\:\:\:\:\:$ -x $\:\:\:\:\:\:$ - $\:\:\:\:\:\:\:\:\:\:\:\:$ +x $\:\:\:\:\:\:\:\:\:\:\:\:$ +x
E $\:\:\:$ 0.043 - x $\:\:\:\:\:\:$ - $\:\:\:\:\:\:\:\:\:\:\:\:$ x $\:\:\:\:\:\:\:\:\:\:\:\:$ x

ka = $1.8 * 10^{-5}$

$1.8 * 10^{-5}$ = $\frac{x^{2}}{0.043}$
$x = 8.8 * 10^{-4} M$
pH = $-log[8.8 * 10^{-4}]$
= $3.05$Is this the right approach?

 

[Saitama]

Your approach is correct but the concentration of $C_2H_3O_2^-$ at E is not x, it is x+something, can you figure out the "something" part? (Think about the reaction of NaOH with the given acid).

What is other approach you talk about? To my knowledge, you have a buffer after reaction of NaOH with $HC_2H_3O_2$ so you might have seen the online resources using Henderson–Hasselbalch equation but that is actually obtained from the procedure you follow, there is really no need to look it up.

 

[Ace.]

Your approach is correct but the concentration of $C_2H_3O_2^-$ at E is not x, it is x+something, can you figure out the "something" part? (Think about the reaction of NaOH with the given acid).

What is other approach you talk about? To my knowledge, you have a buffer after reaction of NaOH with $HC_2H_3O_2$ so you might have seen the online resources using Henderson–Hasselbalch equation but that is actually obtained from the procedure you follow, there is really no need to look it up.

Thanks for the reply. The other approach is where they use a mole chart
like this

I just want to know the situations of when to use a mole chart. This had been confusing me for weeks haha.

 

[Saitama]

I just want to know the situations of when to use a mole chart. This had been confusing me for weeks haha.

You use a mole chart in every situation as you did in your attempt. Did you solve the problem?

 

[epenguin]

From moles and volume you have calculated the molarity of HAc. But you haven't calculated the molarity of Ac^- that I can see.

Write out the formula for K_a and you will see you you don't need to calculate either molarity. [HAc] and [Ac^-] enter as a ratio, molarity ratio equals moles ratio. From the K_a formula and what you have, finding [H⁺] is staightforward enough.

 

[Borek]

You can assume neutralization went to completion, use this assumption to calculate [HAc] and [Ac^-], plug into Henderson-Hasselbalch equation. Done.

You can also start with above, assume dissociation went further and use ICE table to find the exact answer.

The difference will be in most cases negligible, so the latter step can be safely ignored.

These are about buffer pH, but it is exactly the same problem:

http://www.chembuddy.com/?left=buffers&right=composition-calculation

http://www.chembuddy.com/?left=buffers&right=with-ICE-table

 

[Ace.]

Your approach is correct but the concentration of $C_2H_3O_2^-$ at E is not x, it is x+something, can you figure out the "something" part? (Think about the reaction of NaOH with the given acid).

0.029+x?

Oh I see$H_{}C_{2}H_{3}O_{2} + H_{2}O \leftrightharpoons H_{3}O + C_{2}H_{3}O_{2}^{-}$
I $\:\:$ 0.043 $\:\:\:\:\:\:$ - $\:\:\:\:\:\:\:\:\:\:\:\:$ 0 $\:\:\:\:\:\:\:\:\:\:\:\:$ 0.029
C $\:\:\:\:\:\:$ -x $\:\:\:\:\:\:$ - $\:\:\:\:\:\:\:\:\:\:\:\:$ +x $\:\:\:\:\:\:\:\:\:\:\:\:$ +x
E $\:\:\:$ 0.043 - x $\:\:\:\:\:\:$ - $\:\:\:\:\:\:\:\:\:\:\:\:$ x $\:\:\:\:\:\:\:\:\:\:\:\:$ x + 0.029

ka = $1.8 * 10^{-5}$

$1.8 * 10^{-5}$ = $\frac{x * 0.029}{0.043}$
$x = 2.67 * 10^{-5} M$
pH = $-log[2.67 * 10^{-5}]$
= $4.57$

And to verify using henderson equation: $pH = 4.74 + log(\frac{0.029}{0.043}) = 4.57$. Much easier.

Thanks! It makes sense too, I guess my teacher taught me wrong, I'll talk to him about it.

 

[Borek]

Note that x (being 2.67×10^-5) is three orders of magnitude smaller than initial concentrations (both being in the 10^-2 range). Thats why you can ignore it.

It won't be that way for very diluted solutions.