Calculations of Ka,Kc,Kb help

[Guest mnguyen86]

Hi, i have this hwk that i have to do, but i have been stumbling on it for almost a week, and still can't figure out how to do it... well.. if anyone can help, it will be really appreciated.....

 

1)the Ka of benzoic acid(HC7H502) is 6.28x10^-5. What is the pH of a 0.15M solution of the sodium salt of benzoic acid(Na+C7H5O2-)?

 

2)Ammonium carbamate, a crystalline solidundergoes decomposition at 450 degrees C in a closed vessel. The total pressure in the vessel at equilibrium is 0.843 atm. Calculate the value of Kp for the equilibrium.

NH2CO2NH4(g) <-----------> 2NH3(g)+CO2(g)

 

3)A buffer solution is prepared from 0.250 moles of acetic acid(AcOH), pKa = 4.76, and 0.400 moles of sodium acetate(NaOAc). What is the pH of this solution?

 

4)You have 500 ml of a buffer solution of 0.100Mof acetic acid and 0.100 M sodium acetate. Ka = 1.8x10^-5 for acetic acid. Tot his buffer was added 25ml of 0.100 M NaOH. What was the starting pH and the final pH?

 

5)A 0.100M solution of a moderately strong monoprotic acid in water has a pH = 1.88. Calculate the value of Ka(use no simplifying assumptions in this case).

 

Thank you all for your time!

[greg1917]

1) the benzoate ion will act as a base so you need the Kb value for it.

 

Kb = Kw / Ka

= 1.59e-10

 

Now u can do the calculation.

[OH-] = root( c Kb)

= 4.89e-6

[H+] = Kw/ [OH-]

= 2.046e-9

 

pH = -log [H+]

= 8.69

 

2) Are you leaving out any information? Im sure you need the partial pressure of each gas to do this question, not just the total pressure. ammonium carbamate sublimes at...... something close to 60C i think.

 

3) [H+] = Ka ([acid]/[salt])

= 1.74e-5 x (0.25/0.4)

= 4.96

 

4) Apologies, I forget how to do these types of questions. anyone willing ti lend a hand?

 

5) No simplifying assumptions? Well, the way I did it was

[H+] = 10e-(pH)

 

so [H+] = 0.0132

 

If [H+] = root (c ka)

 

then we eventually get a Ka figure of 1.74e-4

 

Can someone else tell me if Im right? Im not too confident about the last one...

[Dudde]

AGH!!

I forgot all about this thread...sorry...umm...I'll check this in tomorrow^_^ how about that!? not sure I'll know it all though...blike or faf? ed?O_o anyone?

 

I'll be back:cool:

[Bsun]

Originally posted by mnguyen86

4)You have 500 ml of a buffer solution of 0.100Mof acetic acid and 0.100 M sodium acetate. Ka = 1.8x10^-5 for acetic acid. Tot his buffer was added 25ml of 0.100 M NaOH. What was the starting pH and the final pH?

 

 

4). The key to remember here is that the strongest acid in appreciable concentration will always react with the strongest base in appreciable concentration.

 

In this case you have a very strong base OH, reacting with the strongest acid in appreciable concentration in this solution, which is HAc(acetic acid) in this fashion:

 

HAc + OH ---> H20 + Ac

 

This reaction goes (practically speaking) to 100%. OH kills any HAc present and forms a water molecule and acetate ion.

 

So work out the stoicheometry

(500*.1 - 25*.1)= mmol HAc

(500*.1 + 25*.1)= mmol Ac-

 

Setup your K expression

K(HAc)= [Ac-]*[H30+] / [HAc]

 

Should be trivial from this point forward.

[Dudde]

d'oh!

...I'm glad we have smart people here...

[wolfson]

I didnt even see this question on the 23rd, must have had my blinkers on.

[wolfson]

And greg as far as I can see 5 is the correct calculation.

[chemistry]

1)the Ka of benzoic acid(HC7H502) is 6.28x10^-5. What is the pH of a 0.15M solution of the sodium salt of benzoic acid(Na+C7H5O2-)?

 

From Ka, find Kb, find pOH then find pH.

 

2)Ammonium carbamate, a crystalline solidundergoes decomposition at 450 degrees C in a closed vessel. The total pressure in the vessel at equilibrium is 0.843 atm. Calculate the value of Kp for the equilibrium.

 

The key here is that the reactant is a solid and thus it is not incorporated in to Kp; in Kp=p of products only. You should know how to do it now. Be sure to pay attention to exponents.

 

3)A buffer solution is prepared from 0.250 moles of acetic acid(AcOH), pKa = 4.76, and 0.400 moles of sodium acetate(NaOAc). What is the pH of this solution?

 

Use the henderson-hasselbach equation.

 

 

 

4)You have 500 ml of a buffer solution of 0.100Mof acetic acid and 0.100 M sodium acetate. Ka = 1.8x10^-5 for acetic acid. Tot his buffer was added 25ml of 0.100 M NaOH. What was the starting pH and the final pH?

 

The hydroxide anion reacts with acetic acid thus decreasing its concentration relative to the base; this is the only significant change. Thus account for the decreased concentration of acetic acid.

 

5)A 0.100M solution of a moderately strong monoprotic acid in water has a pH = 1.88. Calculate the value of Ka(use no simplifying assumptions in this case).

 

Not sure what is meant by the last statement. If it means what I think it means:

 

you know the concentration of hydronium ions in pure water [H+]. Thus pH=-log[H+ of water + H+ of the acid]. Solve for the latter. You should know the rest.

[wolfson]

Yes Chemistry your right just as Greg was!!

[chemistry]

Number 2 was unsolved. I guess I should have attempted this problem only.

[winesmith]

isnt there a buffer created in problem #1? , weak acid and the salt of the weak acid,,, theres a buffer right?,,, so you need to find the initial ph of the acid, then use the HH(Henderson Hasselbach eqt.) at equilibrium after everything has reacted.

 

nevermind, i thought it said the salt was in a solution of benzoic acid. Then you would have a buffer.

[hermanntrude]

thread moved to homework help where it should have been in the first place. Also please note that even when it's NOT in the right place, a thread asking for help with a project or homework or anything similar should NOT be answered immediately with specific detailed answers Give some clues, perhaps refer the asker to a useful website or a textbook chapter, for instance, use some keywords.