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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: jmxwell on October 29, 2020, 10:27:35 AM

Title: Ph question, I'm stuck
Post by: jmxwell on October 29, 2020, 10:27:35 AM

What is the pH of a solution resulting from the addition of 0.2 moles of lactic acid in 1L of distilled water? pKa = 3.85. What will be the pH after adding 0.1 M NaOH? And after adding an additional 0.1 moles of NaOH? And after adding 100 mL of water?
My thoughts:
M= n/V = 0.2/1 = 0.2 mol/L
pH of weak acids ⇒ pH = ½ * (pKa) - ½ * ([[acid])
pKa = 3.85
pH = ½ *(3.85) - ½ *log (0.2) = 1.925 + 0.35 = 2.27
After add NaOH
pH = 3.85 + log [(0.2+0.1/0.2-0.1)] ⇒ ph = 3.85 + 0.477 = 4.327

I would love to know if Im' doing well and how can I find last ph after 100 mL H20, Im' stuck with no toughts on it.

Title: Re: Ph question, I'm stuck
Post by: AWK on October 29, 2020, 10:37:13 AM

For the second part do correct  stoichiometry of neutralization.

Title: Re: Ph question, I'm stuck
Post by: jmxwell on October 29, 2020, 10:45:46 AM

But I can't understand, How does ph will decrease if he/she is adding a strong base to solution?
I can't figure out where is my error

Title: Re: Ph question, I'm stuck
Post by: AWK on October 29, 2020, 10:55:47 AM

And if you add alkali to the water, the pH does not rise?

Title: Re: Ph question, I'm stuck
Post by: jmxwell on October 29, 2020, 11:02:29 AM

Will increase, but you told my calculous are wrong in second part. ???

Title: Re: Ph question, I'm stuck
Post by: AWK on October 29, 2020, 11:06:50 AM

Because your calculation are wrong.

See: https://www.chemicalforums.com/index.php?topic=105821.0

Title: Re: Ph question, I'm stuck
Post by: jmxwell on October 29, 2020, 11:27:27 AM

CH3CHOHCOOH + NaOH -> CH3CHOHCOONa + H2O

pH = 3.85 + log [(0.1/0.2)] ⇒ ph = 3.85 - 0.30 = 3.55

Title: Re: Ph question, I'm stuck
Post by: AWK on October 29, 2020, 11:31:25 AM

Can't do simple subtraction? After all, the acid reacts with the base.

Title: Re: Ph question, I'm stuck
Post by: jmxwell on October 29, 2020, 11:36:39 AM

pH = 3.85 + log [(0,2-0.1/0.1+0.2)] ⇒ ph = 3.85 - 0.48 = 3.37

Title: Re: Ph question, I'm stuck
Post by: AWK on October 29, 2020, 11:39:15 AM

Quote from: jmxwell on October 29, 2020, 11:36:39 AM

pH = 3.85 + log [(0,2-0.1)/0.1+0.2)] ⇒ ph = 3.85 - 0.48 = 3.37

?

Title: Re: Ph question, I'm stuck
Post by: jmxwell on October 29, 2020, 11:45:41 AM

I thought the added concentration would be always some amount - the concentration of the acid or base/ concentration of add compound plus the remain of the acid.

pH = 3.85 + log [(0,2-0.1/0.1)] = 3.85 + 0 = 3.85

So occurs a neutralization, and like the example you gave me ph= pka because its a strong base

Title: Re: Ph question, I'm stuck
Post by: AWK on October 29, 2020, 11:50:20 AM

Quote from: jmxwell on October 29, 2020, 11:45:41 AM

I thought the added concentration would be always some amount - the concentration of the acid or base/ concentration of add compound plus the remain of the acid.

pH = 3.85 + log [(0,2-0.1)/0.1)] = 3.85 + 0 = 3.85

So occurs a neutralization, and like the example you gave me ph= pka because its a strong base

OK

Title: Re: Ph question, I'm stuck
Post by: jmxwell on October 29, 2020, 11:52:40 AM

But I don't know how to start thinking about adding 100 mL, how to start thinking about this question?
C(a)* V(a) = C(b)*V(b)?

Title: Re: Ph question, I'm stuck
Post by: AWK on October 29, 2020, 11:56:17 AM

After adding water to the buffer solution, you change the concentration in the numerator and denominator proportionally.

Title: Re: Ph question, I'm stuck
Post by: jmxwell on October 29, 2020, 12:35:07 PM

I have no idea, because there isn't concentration of water

Title: Re: Ph question, I'm stuck
Post by: jmxwell on October 29, 2020, 12:48:49 PM

[H3O+] = Ka(Ca - [H3O+])/(Cb +[H3O+]
[H3O+]^2 + (Cb +Ka)[H3O+] - KaCa = 0
x^2 + (0.2+3.85)x -(3.85.0.1) = 0
x^2 + 4.05x - 0.385=0
x=0.09
ph = -log[H+] = -log(0.09) = 1.04
ph = 3.85+1.04 =4.89