You are Here: Home

Acids and Bases ... watch

1. I'm stuck on parts of this question ... I have a graph showing the pH changes when 0.12 mol dm-3 NaOH is added to 25 cm3 of a solution of a weak monoprotic acid, HA. (I've attached a file showing a similar graph to the one i have been given to illustrate the question a little clearer)

i) Use the graph to calculate the initial concentration of the acid HA. (2)

At the point where no alkali has been added, at 0 cm3, my pH value of acid is 1.6 ... (is this the right point at which i should be taking the pH reading?) ... from the 1.6, I then get ...

pH = - log10 [H+] giving me a [H+] concentration of 0.0251 mol dm-3.

I'm not sure where to go from here ... is there another step, or do i scale the value out of 25cm3 somehow?

ii) Write the expression for the dissociation constant, Ka, of the weak acid HA. (1)

Ka = [H+] [A-] / [HA]

iii) Determine the volume of sodium hydroxide added when [HA] = [A-] and use the graph to determine the pH at this point. (2)

Is this simply the half equivalence point? 11.8 cm3 of alkali is added to reach the equivalence point, so it would be the value of pH at 5.9?

iv) Use the answers to part 2 and 3 to determine the value of Ka for the acid HA (4)

I don't understand which values, how the volume or pH values can be substituted into the equation for the dissociation constant ...

Any help and tips much appreciated xx
Attached Images

2. are you sure it was a weak acid? that curve is for a stong acid/strong base.
3. (Original post by vinny221)
are you sure it was a weak acid? that curve is for a stong acid/strong base.
That graph was just to illustrate the general shape of the curve, and show the axis ... the question does say weak acid, and displays a curve *similar* to that I posted ... i do apologise however, i wasn't too careful in finding a perfect graph regarding the strength of the acid!
4. (Original post by h1nna)
I'm stuck on parts of this question ... I have a graph showing the pH changes when 0.12 mol dm-3 NaOH is added to 25 cm3 of a solution of a weak monoprotic acid, HA. (I've attached a file showing a similar graph to the one i have been given to illustrate the question a little clearer)

i) Use the graph to calculate the initial concentration of the acid HA. (2)

At the point where no alkali has been added, at 0 cm3, my pH value of acid is 1.6 ... (is this the right point at which i should be taking the pH reading?) ... from the 1.6, I then get ...

pH = - log10 [H+] giving me a [H+] concentration of 0.0251 mol dm-3.

I'm not sure where to go from here ... is there another step, or do i scale the value out of 25cm3 somehow?

ii) Write the expression for the dissociation constant, Ka, of the weak acid HA. (1)

Ka = [H+] [A-] / [HA]

iii) Determine the volume of sodium hydroxide added when [HA] = [A-] and use the graph to determine the pH at this point. (2)

Is this simply the half equivalence point? 11.8 cm3 of alkali is added to reach the equivalence point, so it would be the value of pH at 5.9?

iv) Use the answers to part 2 and 3 to determine the value of Ka for the acid HA (4)

I don't understand which values, how the volume or pH values can be substituted into the equation for the dissociation constant ...

Any help and tips much appreciated xx

Ka = [H+] [A-] / [HA]

Using [HA] = [A-]

Ka = [H+]2 / [HA]

[HA] = [H+]2 / Ka

So, all you need to do is substitute the numbers into that equation

5. iii) i don't really know if you have the right idea without the graph... but the pH value does sound a little high so if i were you i'd double check what the meaning to half equivalence is to be sure

iv) Ka = [H+] [A-] / [HA]

At half equivalence, [HA] = [A-]

Therefore,

Ka = [H+]

So, use...

[H+] = 10-pH

And then use Ka = [H+] to work out the Ka
6. I) pH doesn't matter, what matters is amount of base added to reach equivalence point. Write neutralization reaction and calculate amount of acid from simple stoichiometry.

iv) that's where so called Henderson-Hasselbalch comes in handy - as Revenged wrote [HA] and [A-] cancels out, thus pKa = pH (compare this to H-H equation and try to understand why!).

Best,
Borek
--
7. when pKa and pH are equal you are halfway thru the buffering part of the curve, no?
8. Yes.

Best,
Borek
--
9. is this ocr?
10. Henderson-Hasselbalch equation, is beyond A-Level, just did last week and am doing BSc Year 1

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: November 20, 2005
Today on TSR

University open days

• Southampton Solent University
Sun, 18 Nov '18
Wed, 21 Nov '18
• Buckinghamshire New University