The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

This discussion is now closed.

Check out other Related discussions

can someone explain these damn questions

If 4.8g of carboxylic acid (RMM 122), PKA 6.7 is dissolved in 50mL Water, calculate the PH of the resultant solution.


Calculate the PH of an aqeuos 0.32M solution of the sodium salt of the carboxylic acid (pka 3.7)

can someone take me through them
Reply 1
Avatar for JayEm
JayEm
14
1. First calculate the concentration of the solution, to do this you need to know the number of moles of the carboxylic acid:

moles = mass/Mr
moles = 4.8 / 122 = 3.934426 x 10-2 mol

Concentration of solution:
use M=(1000n)/V
M = (1000 x 3.934426 x 10-2) / 50
= 7.8688 x 10-1M

Next, to find the pH, use the equation Ka = [H+]2/[HA]

You need to convert the pKa value to its Ka value before you can plug in the values.

To convert pKa of 6.7, you take the antilog or 10-6.7 = 1.9953 x 10-7

So now you can rearrange the above equation to obtain the [H+] value.

[H+] = (1.9953 x 10-7 x 7.8688 x 10-1)1/2
= 3.9624 x 10-4

pH = -log10(3.9624 x 10-4)
= 3.4

You're on your own with the next one. :p:
Reply 2
Avatar for vVPiKaVv
vVPiKaVv
1
For these type of questions, a general approach is as follows:

Identify what is in the solution at any instant in time, then apply the formula/method to solve it.
If it has strong acid/base, use pH = -lg [H+] and pOH = -lg [OH-]
If it has weak acid/base, use the approximate method (not sure if you have to check whether the approximate method is suitable), and then use pH = -lg [H+] and pOH = -lg [OH-].
If it has buffer solution, use pH = pKa + lg {[salt]/[weak acid]} and pOH = pKb + lg {[salt]/[weak base]}.

Also remember that pH = 14 - pOH, and
at 25°C for a acid-conjugate base pair or base-conjugate acid pair, Ka x Kb = Kw.

For example, the first question, you have weak acid and water in the solution. So, use approximate method and then calculate pH.
Reply 3
Avatar for reaney
reaney
OP
thanks to both of you but I have some more questions. The teachers are not helping us hence I am struggling a bit.

For the question which says calculate the PH of an aqeous 0.32M solution of sodium salt of a carboxylic acid (pka 3.7) would I use the ka equation again.
edit: I had a go at this question, got ph to be 2.097, this is going to sound daft but if I wanted to round to 2 decimal places would it be 2.01?
no, it would be 2,10
Reply 5
Avatar for reaney
reaney
OP
thanks for that charco, what about this question

find the ph of a mixture of 20ml 0.5m CH3COOH and 25ml 0.5M Naoh
The NaOH reacts with all of the ethanoic acid so the final mixture only contiains OH ions which affect the pH.

Calculate the moles of NaOH remaining. You have the final volume so you can calculate the molarity of the NaOH and from there the pH
Reply 7
Avatar for reaney
reaney
OP
ok thank you. I have a few more questions, sorry bout this but we havent been taught even the basics,.

Identify the conguate acid/base pairs

1. HN03 +CH30H =NO3- + CH30H2+
2. H20+ CH3COO- = OH- + CH3COOH
3. HCO3- +NH3= NH2- + H2C03
go to the ib chem .com website ..... navigate to acids and bases and all your doubts will be resolved
Reply 9
Avatar for reaney
reaney
OP
charco
The NaOH reacts with all of the ethanoic acid so the final mixture only contiains OH ions which affect the pH.

Calculate the moles of NaOH remaining. You have the final volume so you can calculate the molarity of the NaOH and from there the pH

thanks for the website. does the final volume have to be in ml or l?
when calculating molarity it is in moles per LITRE so it must be in litres

Related discussions

Latest

Trending

Trending

Articles for you