Help with PH Question (1 Viewer)

[Pinioned]

A 250 mL sample of rainwater is titrated using 0.0095 mol L-1
sodium hydroxide solution.
If 8.3 mL of the sodium hydroxide solution was required to reach the endpoint, what was the pH
of the rainwater?
(A) 0.54
(B) 3.50
(C) 4.50
(D) 5.40

Apparently the answer was B, but i got 3.5 in my working, and did -log10[H+], not sure what i did wrong

 

[Ekman]

A 250 mL sample of rainwater is titrated using 0.0095 mol L-1
sodium hydroxide solution.
If 8.3 mL of the sodium hydroxide solution was required to reach the endpoint, what was the pH
of the rainwater?
(A) 0.54
(B) 3.50
(C) 4.50
(D) 5.40

Apparently the answer was B, but i got 3.5 in my working, and did -log10[H+], not sure what i did wrong

The answer you got is B, hence you got it correct

 

[Pinioned]

Shit, no as in i got A****

 

[DepressedPenguino]

Shit, no as in i got A****

Hmm u mean the answer is A or u got A? you said u got 3.5 as your ans and that is B.

 

[Occupied]

A 250 mL sample of rainwater is titrated using 0.0095 mol L-1
sodium hydroxide solution.
If 8.3 mL of the sodium hydroxide solution was required to reach the endpoint, what was the pH
of the rainwater?
(A) 0.54
(B) 3.50
(C) 4.50
(D) 5.40

Apparently the answer was B, but i got 3.5 in my working, and did -log10[H+], not sure what i did wrong

Are you sure you didn't mean pOH of the rainwater = 3.5 and not pH of rainwater is 3.5?

Assuming ^ is true:
-log10 [ (0.0095 * 0.0083)/0.25 ] = 3.501138311 = 3.5 =pOH

Otherwise if it's to find the pH then:

14+ log10 [ (0.0095 * 0.0083)/0.25 ] = 10.498886169 = pH

 

[Ekman]

Actually one assumption has been made with this question. It has assumed that rainwater contains only a monoprotic acid and not a mixture of di or tri protic acids.

 

[Occupied]

Actually one assumption has been made with this question. It has assumed that rainwater contains only a monoprotic acid and not a mixture of di or tri protic acids.

? Still waiting for OP to respond so we can find out whether he typoed 'to find ph' instead of 'to find pOH'.
if it's to find pH I can see where you're coming from, but my calculations are only.giving me 3.5 when I find pOH, and if it's to find pOH I don't understand how proticness of an unknown acid affects the question.

 

[Pinioned]

Yeah sorry aha, i have seemed to calculate just pH instead of pOH, forgot to do 14-3.5, thankyou very much

 

[Ekman]

Yeah sorry aha, i have seemed to calculate just pH instead of pOH, forgot to do 14-3.5, thankyou very much

But 14-3.5 wont give any answer that the mc gives...

 

[zhertec]

Why is everyone saying the POH is 3.5? the pH is 3.5...
Because when you are calculating the titration question:
C(a) x V(a) = C(b) x V(b)

And you calculate that C(a) = (0.0083*0.0095)/0.250
^that is the concentration of the acid, not the base....and when you -log(base 10) (H+ conc.) that is obtaining the pH of the solution not POH.

Hence B (assuming the carbonic acid used only ionises once).

 

[Fiction]

Why is everyone saying the POH is 3.5? the pH is 3.5...
Because when you are calculating the titration question:
C(a) x V(a) = C(b) x V(b)

And you calculate that C(a) = (0.0083*0.0095)/0.250
^that is the concentration of the acid, not the base....and when you -log(base 10) (H+ conc.) that is obtaining the pH of the solution not POH.

Hence B (assuming the carbonic acid used only ionises once).

Am I missing something? Where are you getting carbonic acid from?

 

[zhertec]

Am I missing something? Where are you getting carbonic acid from?

That was assuming the rainwater was pure. Regardless, it does not change the value of the pH since we are assuming the acid is of monoprotic nature.

 

[Fiction]

That was assuming the rainwater was pure. Regardless, it does not change the value of the pH since we are assuming the acid is of monoprotic nature.

o.o How are we determining that the acid is of a monoprotic nature? How could you make a judgement that this is pure rainwater...? And rainwater has carbonic acid?

I would calculate it by calculating pOH rather than take all of ^ into account tbh

Also, I don't get what you mean by

And you calculate that C(a) = (0.0083*0.0095)/0.250
^that is the concentration of the acid, not the base....and when you -log(base 10) (H+ conc.) that is obtaining the pH of the solution not POH.

Like you said, that's finding the concentration of the acid, but you're finding the concentration of the acid by using values assigned to the base? I don't comprehend :/

 

[Ekman]

First of all, you have to look at the nature of the experiment. Its a titration between a standardised base and an unknown acid. By calculating the pOH of the NaOH, there is no need for the titration to go on, as any person can use the formula's c=n/v and pOH=-log[OH-]. However we are measuring the pH of rain water, hence the assumption is made that the acid is a monoprotic acid. So you have to construct a chemical equation like this:
H+ + NaOH -> H2O + Na+. So you calculate the H+ concentration via mole ratios, and use the pH=-log[H+] formula to get 3.5.

Otherwise if you calculate the pOH of NaOH, you will get 3.5, and using the formula 14=pH + pOH, pH=14-3.5=11.5, hence you are saying that the rainwater is basic, so you titrated something that was basic with another basic substance... see the mistake?

 

[zhertec]

o.o How are we determining that the acid is of a monoprotic nature? How could you make a judgement that this is pure rainwater...? And rainwater has carbonic acid?

I would calculate it by calculating pOH rather than take all of ^ into account tbh

Also, I don't get what you mean by

Like you said, that's finding the concentration of the acid, but you're finding the concentration of the acid by using values assigned to the base? I don't comprehend :/

The judgement of monoprotic was determined by the community. I just assumed it was carbonic acid, although this may initially cause the change in ratios, the fact that people say it is monoprotic ensures that it only ionises once, and essentially does not affect the ratio of acid to base.

To my understanding pOH = 14 - pH.
But the issue with finding the pOH is that we do not have to the total volume of the solution. The titration equation was used instead of finding the pOH of the solution as it takes into account the concentration of the acid/base after they have reacted together. In order to find the pOH of the solution we need to know the concentration of the base in mols/L. The 0.0095 mol/L(-1) value is given as the concentration prior being titrated with the rainwater. The new concentration of the base is now unknown in the solution.

The values that are assigned to the base help us to find the concentration of the acid, since the definition of equivalence point is the point where stoichiometrically, the moles of acid equal to the moles of base. Hence the n(acid) = n(base), we can therefore expand this to C(a) x V(a) = C(b) x V(b). Then just by subbing in values and making C(a) the subject we get the thing I wrote above.

 

[someth1ng]

Why we assuming the acid is monoprotic? Carbonic acid is the main acid responsible for the lowered pH of rainwater.

Anyway, you can't determine the pH before just from titration. Take for example, 1 M HCl vs 1 M citric acid, you would need 3x more citric acid to titrate but the pH would probably be higher since it's a much weaker acid than HCl so really, pH isn't all that relevant.