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April 16, 2014

April 16, 2014

Posted by **Danielle** on Monday, April 30, 2007 at 3:54pm.

Solution A is 50.00 mL of a 0.100 M solution of the weak monoprotic acid HX.

Solution B is a 0.0500 M solution of a salt NaX. It has a pH of 10.02.

Solution C is made by adding 15.00 mL of .250 M KOH to solution A.

How much do you know what to do? I don't want to duplicate that which you already know. For starters, however,

solution A is a weak acid. I assume you know how to determine the pH of this solution. The Ka for HX is determined by working solution B FIRST.

Post any work if you need further assistance AND explain exactly what yu don't understand about each part of the problem. I would work solution B first, followed by solution A, then C. I must say someone is making you work for the ton of money.

I have determined the Ka for solution B, now how do I figure out solution A?

HX ==> H^+ + X^-

Initial concentrations.

HX = 0.1

H^+ = 0

X^- = 0

What happens when HX ionizes.

y molar of HX ionizes producing y molar H^+ and y molar X^-

At equilbrium. What are the concentrations.

(H^+) = y

(X^-) = y

(HX) = 0.1 - y

Ka = (H^+)(X^-)/(HX)

Plug in y and 0.10-y in appropriate spots and solve for H^+, then convert to pH.

So i use the Ka that I got for solution B?

Yes. The acid in solution A is HX, solution B is the SALT, NaX, therefore, you determined Ka for HX when you did solution B.

Then to figure out C I subtracted the moles of A - moles of C. I then used the difference in the equilibrium equation. For the Ka I used the Ka value of B again so the equation was Ka of B = X^2 / .49625-x. .49625 is the difference. I then found the pH using X. Is this right?

I don't understand what you mean by

"the diffeerence in the equilibrium equation." I know the x^2 is not right. What did you get for Ka before we get too far out.

I got 9.54992586 for the Ka value. the difference is the difference in moles which I then used for my starting rate for the NaOH in an ICE table

9.54992586 x 10^-11

I might believe you obtained 9.5499E-11 the H^+ of solution B but that isn't Ka.

That is the Ka of solution B. I then used Ka = (H^+)(X^-) / HX with HX being .1-x and the ions being x. I then solved for x.

I then took the value for X and plugged it into the pH equation and got a pH of 4.51 for solution A.

For solution B. 0.05 M NaX has pH = 10.02.

X^- + HOH ==> HX + OH^-

Kb = (HX)(OH^-)/(X^-) = Kw/Ka

pH = -log(H^+) = 10.02

(H^+) = 9.55E-11

(OH^-) = 1.047E-4

(HX) = 1.047E-4

(OH^-) = 1.047E-4

(X^-) = 0.05 - 1.047E-4 = 0.0499 (or 0.05)

[(1.047E-4)^2/0.05]= Kw/Ka

Kw = 1E=14

Ka = ??

Check my thinking.

I don't know

pH = -log Ka

10.02 = -log Ka

this then gave me the Ka of solution B, I think.

No it doesn't.

pH is not -log Ka.

pH = -log(H^+).

So plug in 10.02 for pH and

(H^+) = 9.55E-11.

Then you must use (H^+) = 9.55E-11 and calculate Ka for HX which is what I did above.

I worked out all but Ka.

Ka = 4.56E-8

Due to the Henderson-Hasselbalch Equation, pH = pKa that is what I used for a lab I just did.

If what I am saying is wrong what do I do now with the H+ value?

The HH equation is

pH = pKa

That is where you are going wrong.

pH is not pKa, it is pKa + all that other stuff. And since solution B is a salt you a concentration for the base (the X^-) but you don't have a concentration for the acid (HX) because there is no acid in solution B. The HH equation is for buffers. It is of no value for solutions of plain salts. There must be an acid AND a base to be a buffer. I don't know what lab you did BUT I expect you titrated an acid with a base and determined the pKa from the HH equation. But if you will look in your lab notes, you will find that you took the value of pH

pH = pKa + log [(base)/(acid)]

pH = pKa + at the half way point, the (base)=(acid) so base/acid = 1 and log 1 = 0 and pH = pKa. But that is ONLY true at the half way point. Not other point. And of course, from what I said above, the HH equation is of no value UNLESS you have a buffer. A salt like NaX is not a buffer. You must use the hydrolysis equation to solve for pH.

So then I us Ka = (H+)(X-) / (HX)

and get X=6.752777206E-5 and then take the negative log to get a pH of 4.170517579

Not for solution B.

The base (X^-) hydrolyzes as follows:

X^- + HOH ==> HX + OH^-

Look back a few posts back and you will see where I wrote that equation and wrote the equilibrium constant as

Kb = [(HX)(OH^-)/(X^-)] = Kw/Ka

You know Kw, (HX)=(OH^-) = 1.047E-4 and (X^-) = 0.05- 1.047E-4.

Solve for Ka = 4.56E-8

So it is the Ka for solution A. I am so confused. We never learned this!! Now waht do I do with this Ka value?

Let's start over.

Step 1. Use solution B and pH = 10.02 to calculate Ka. You can find at least two of my posts that go through that. I found Ka = 4.56E-8

Step 2. go to solution A. It is a weak monoprotic acid, HX.

Knowing Ka, determine the pH of the 0.1M solution.

HX ==> H^+ + X^-

Ka = (H^+)(X^-)/(HX)

H is y, X is y and HX is 0.1-y.

Solve for H. I found 6.75E-5 for H and pH = 4.17

Step 3. Go to solution C.

You may use the HH equation here.

pH = pKa + log (base)/(acid)

pH = 7.34 = (3.75/V)/(1.25/V).

pH = 7.82.

I hope this helps. We have so much here that I am getting confused, too.

Let me know if you have any problems. If you want to ask about anything, let's keep it to what I have in THIS response. Please copy it with a new post and I can answer with that data before me. Again, I hope this helps. Thanks for using Jiskha.

I made a typo here.

Step 3. Go to solution C.

You may use the HH equation here.

pH = pKa + log (base)/(acid)

pH = 7.34 = (3.75/V)/(1.25/V).

pH = 7.82.

- bvxo abged -
**bvxo abged**, Sunday, February 1, 2009 at 9:23pmtvhow adxnwrv xpnja fpemnkgwc ikvwtf qdugkr qxmwcb

- Chemistry -
**ML**, Thursday, May 14, 2009 at 8:02pmHint:

Solution B is not used to solve this type of problem because it only involves solutions A and C.

First, you would have to convert 50 milliliters to liters: (50 milliliters of .100M HX /1) X (1 liter of HX/ 1,000 liters of .100M HX) X ( .100 moles of HX/1 liter of HX) = .005 or 5 X 10^-3.

After you get the moles of HX, convert it to pH = -log(.005) = 2.301

For solution C, you follow the same procedure as the one demonstrated above.

Since KOH and HX is a 1:1 mole ratio, you can subtract the moles of Soln. C from Soln. A.

The answer represents the pOH. In order to find the pH, you must subtract the known value( pOH ) from 14.

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