[jmxwell]

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.

[AWK]

For the second part do correct  stoichiometry of neutralization.

[jmxwell]

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

[AWK]

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

[jmxwell]

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

[AWK]

Because your calculation are wrong.

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

[jmxwell]

CH3CHOHCOOH + NaOH -> CH3CHOHCOONa + H2O

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

[AWK]

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

[jmxwell]

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

[AWK]

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

?

[jmxwell]

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

[AWK]

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

[jmxwell]

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)?

[AWK]

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

[jmxwell]

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