Thursday

September 29, 2016
Posted by **Dan** on Monday, April 30, 2007 at 8:37pm.

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.

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 = ??

I understand all of this but for some reason I am not getting the same Ka value you got of Ka = 4.56E-8. I am using (1.047x10^-4)^2 / .05 = (1.0x10^-14) / Ka

Sorry I'm a little slow. I had to eat supper and do my 2 mile walk. I worked it again and came up with 4.56E-8 again. If you don't find your error, post your work and I'll nail it for you.

No, problem

(1.047x10^-4)^2 = 1.096478196x10^-8

1.096478196x10^-8 / .05 = 2.192956392x10^-7

2.192956392x10^-7 / 1.0x10^-14 = 21929563.92

HELP!!

No, problem

(1.047x10^-4)^2 = 1.096478196x10^-8

1.096478196x10^-8 / .05 = 2.192956392x10^-7

2.19 x 10^-7 = Kw/Ka

2.19 x 10^-7 = 1x10^-14/Ka

Ka = 1x10^-14/2.19x10^-7

Ka = 4.56E-8

2.192956392x10^-7 / 1.0x10^-14 = 21929563.92

HELP!!

I have one final question. Where did the numbers for solution C come from? I am a bit confused. Is there another way to do it?

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).

should be pH = 7.34 + (3.75/V/1.25/V) = 7.82. I typed = instead of +.

pH = 7.82.

HX + KOH ==> KX + HOH

mols HX = 0.05 L x 0.1 M = 0.005 mols

(I used 5 millimols I believe)

mols KOH = 0.015 L x 0.25 M = 0.00375 mols.

(I think I used 3.75 millimols).

So 0.00375 mols KX is formed, all of the KOH is used, and the HX remaining is 0.005 - 0.00375 = 0.00125 (and I used 1.25 millimols for this).

The volume is 15 mL + 50 mL = 65 mL or 0.065 L.

Technically, (HX) = 0.00125 mols/0.065 L but I saved myself some work and just wrote 0.00125/V because I know the V will cancel (actually I wrote 1.25 because I used millimols).

Then (X^-) = 0.00375 mols/0.065 L or 0.00375 mols/V (and I used 3.75 millimols here).

If Ka = 4.56E-8, then pKa = 7.34

The HH equation is

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

pH = 7.34 + log (0.00375/V)/(0.00125/V)

pH = 7.34 + log 3.00. (You see why I didn't stick the 65 mL in because the V appears both places and will cancel. I also used millimols of 3.75/1.25 because that's easier than all those zeros). And 3.75/1.25 is the same as 0.00375/0.00125 anyway.

pH = 7.34 + log 3.00

pH = 7.34 + 0.477 = 7.818 which I rounded to 7.82.

Yes, there is another way to do it and not use the HH equation. Just use the Ka expression.

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

H+ = solve for this.

X- = 0.00125 mols/V

HX = 0.00375 mols/V

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

(H^+) = 4.56E-8*(0.00125/0.00375)

(H^+) = 1.52E-8

pH = -log 1.52E-8

pH = 7.82

THANK YOU SO MUCH! YOU ARE MY LIFE SAVER!

Let me say that I made a mistake. The mistake I made was not asking you for Ka when you said you had it and asked what to do next. By the time we finished that session I was confused, too, about which number I had used where. If I had asked for Ka, we would have taken care of that problem RIGHT THEN, instead of waiting until we were near the end. By the time I found Ka was not right we had talked about hydrolysis, Ka, Kb, Kw/Ka, monoprotic weak acids and the HH equation, and it just got to be too much. Although I didn't give you any false directions, I had so many numbers running around in my head that I was tearing my hair out. Anyhway, I hope I helped you. Thanks for using Jiskha, and come again.

- Chemistry -
**share**, Saturday, April 28, 2012 at 10:49pmpH=1/2pKa-1/2(log (C)

first solution pH should be 4.17