[stardancer]

I feel bad that I have to ask here. I try to struggle my way through my work by myself. But I spent several hours on this question last night and I just don't get it! I'm getting really frustrated!

0.10M lactic acid and 0.10M NaOH solution are needed to make 800 ml of a buffer of 3.50. what is the new pH if 0.10g of NaOH are added? if 0.15g of HCl is added?

Any help is appreciated. Thanks in advance!

Star

[Borek]

A little bit ambiguous. I guess it should be read "0.10M lactic acid and 0.10M NaOH solution were used". That means - use Henderson-Hasselbalch equation to calculate initial amounts of acid and base (note: sum of volumes = 0.8L). Then assume whatever was added reacted stoichiometrically with whatever was in the solution.

[shanmac]

I'm working on this problem as well.  I have calculated the ratios of acid and base, but am not sure what you mean by

Then assume whatever was added reacted stoichiometrically with whatever was in the solution.

Lactic acid disassociates as follows:
CH3CH(OH)COOH (aq) + H2O (l) <---------> CH3CH(OH)COO- (aq)+ H2O (l)
It has a Ka of 1.37 x 10-4 and a pKa of 3.86.

pH = pKa + log [conjugate base]
            [acid]

3.50 = 3.86 + log [NaOH]
           [CH3CH(OH)COOH]

-0.36 = log [NaOH]              
            [CH3CH(OH)COOH]

10-0.36 = 0.44

So a mole ratio of acid:base would be 0.44:1. Therefore, 4.4 L of 0.10 M lactic acid would react with 10 L of 0.10 M NaOH to produce 14.4 L of solution. 
14.4 x 0.800 L = 18
4.4 ÷ 18 = 0.244 L
10 ÷ 18 = 0.556 L
To produce an 800mL buffer solution with a pH of 3.50, we would require 244mL of lactic acid and 556 mL of NaOH.

Now, adding in 0.10 g of NaOH is obviously going to increase the pH.  NaOH is 40 g/mol, so .10 g is 4 moles.  Does this mean that the new ratio of acid: base is now 0.44:5?

[shanmac]

Oops.    Really bad math error... I meant 0.10 g NaOH = 2.5 x 10^-3 mol. So would the new ratio of acid:base now be 0.44: 1.0025?

[Sophia7X]

So a mole ratio of acid:base would be 0.44:1. Therefore, 4.4 L of 0.10 M lactic acid would react with 10 L of 0.10 M NaOH to produce 14.4 L of solution.

Is NaOH really the conjugate base of lactic acid? It is a strong base so you can't use it in the Henderson Hasselbalch equation.

Also, double check your equation.

The ratio calculated is not acid: base, it is conjugate base: acid based on the Henderson Hasselbalch equation. If you want acid:base, you're going to have to inverse it. you can also use the buffer equation to get the acid:base ratio, which is basically the Henderson Hasselbalch without the mathematical manipulations. [H+] = Ka*[Weak acid/conjugate base]

When you get a ratio of conjugate base to acid, you can figure out how much you have of the conjugate base vs the acid.

Now, adding in 0.10 g of NaOH is obviously going to increase the pH.  NaOH is 40 g/mol, so .10 g is 4 moles.  Does this mean that the new ratio of acid: base is now 0.44:5?

Adding more NaOH will disturb the buffer. Don't use the previous ratio; do some stoichiometry (NaOH isnt going to sit there and not react with acid) and have another Henderson Hasselbalch equation set up.

[shanmac]

Thanks Sophia7X.
I did catch that reversal error but forgot to update my post.  Does this look better?

pH = pKa + log [conjugate base]
            [acid]

3.50 = 3.86 + log [CH3CH(OH)COO-]
           [CH3CH(OH)COOH]

-0.36 = log [CH3CH(OH)COO-]              
            [CH3CH(OH)COOH]

10-0.36 = 0.44

So a mole ratio of base:acid would be 0.44:1. Therefore, .44 L of 0.10 M NaOH would react with 1 L of 0.10 M lactic acid to produce 1.44 L of solution. 
0.44x + 1x = 800mL
1.44x=800
x= 556.
800mL - 556mL = 244mL
To produce an 800mL buffer solution with a pH of 3.50, we would require 244mL of 0.10 M NaOH and 556 mL of 0.10 M lactic acid.

Mathematically, this appears to work out.  Hope it's right so far!

Now, for the additional 0.10 g of NaOH here's what I came up with.  Am I on the right track?


Lactic acid = C3H6O3
C3H6O3 (aq) + NaOH (s) <-----------------> C3H6O3Na (aq)+ H2O (l)
The pH of the buffer system is 3.5.
The addition of 0.10 g/40.0 g mol^-1 =  2.5 x 10^-3 moles of OH- forces the equilibrium to the right. The OH- will react completely with the H+ from the lactic acid of the buffer solution, forming water, removing hydrogen ions from the solution.

In 800mL of the above buffer solution, there is (0.10 M x 0.556L x 0.800L =) 0.04 moles of NaOH.

If we add 2.5 x 10^-3 moles of NaOH, this then makes 0.0425 moles of NaOH.

pH = pKa + log [conjugate base]
         [acid]

pH = 3.86 + log [(0.0425 x 0.800]
         [0.016]

pH = 4.19

 If 0.10 g of NaOH were added to the buffer solution above, the new pH would be 4.19.

[Borek]

pH = pKa + log [conjugate base]
            [acid]

You can use LaTeX to properly format such equations:

[tex]pH = pK_a + \log \frac{[conjugate\ base]}{[acid]}[/tex]

(see About post formatting... link above the edit field).

So a mole ratio of base:acid would be 0.44:1. Therefore, .44 L of 0.10 M NaOH would react with 1 L of 0.10 M lactic acid to produce 1.44 L of solution.

Close, but no banana. You are either still treating NaOH as a conjugate base, of forgetting that when the conjugate base (Lactate^-) is produced, it means some of the acid (HLactate) is consumed.

[shanmac]

Right.  So is this where I am wrong?

So a mole ratio of salt:acid would be 0.44:1. Therefore, 0.44 L of 0.10 M NaOH would react with 1 L of 0.10 M lactic acid to produce 1.44 L of solution. 

The math is still right, is it not?

[Borek]

No, math is not right. Mixing solutions in the way you suggest will not give expected concentration ratio of the acid and its conjugate base.

[Sophia7X]

Right.  So is this where I am wrong?

So a mole ratio of salt:acid would be 0.44:1. Therefore, 0.44 L of 0.10 M NaOH would react with 1 L of 0.10 M lactic acid to produce 1.44 L of solution. 

The math is still right, is it not?

No, a ratio of 0.44 conjugate base: 1 acid means that there are 0.44 moles of lactate for every 1 mol of lactic acid. convert this ratio into a percentage so you can figure out how many mL of the 800 mL total solution is acid and conjugate base.

[shanmac]

Ugh.  OK.  Got a new textbook, and I think a light went on...

So.  Working with the steps I already completed, I have a ratio of 0.44 moles of lactate (ie: salt) for every 1 mole of lactic acid.

I just realized that this is the first very small step.

Now, if I multiply both of those numbers by 0.10 (because I am working with 0.10 M solutions of both lactic acid and NaOH), I get 0.044 M lactate
                                             -------------------
                                             0.10 M lactic acid.
Then if I multiply both the top and the bottom by 0.800 (because I want to end up with 800mL of solution), this gives me 0.0352 L lactate
                                ---------
                                0.0800 L lactic acid

...which is equivalent to 35.2 mL over 80.0 mL.

In that I needed 35.2 mL of lactic acid to produce the lactate, I need (35.2 mL + 80 mL) a total of 115.2 mL of 0.1 M lactic acid solution.  I will also need 35.2 mL of 0.1 M NaOH solution.  If I top this off with 649.6 mL of water, I have a total of 800mL of buffer solution.

Am I on the right track now???

[Borek]

Yep, you are on the right track now.

Using your numbers I got pH of 3.54, most likely because you have used rounded down ratio of 0.44:1 - you shouldn't round down intermediate values (round them down for reporting, but use at least 2 or 3 guard digits for calculations).