[영원한 사랑]

1. A buffer solution contains 0.10 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 L
a. what is the pH of this buffer?
b. what is the pH of the buffer after the addition of 0.02 mol of KOH?


2. A 20 mL sample of 0.200 M HBr solution is titrated with 0.200 M NaOH solution. Calculate the PH of the solution after the following volumes of base have been added:
a. 15 mL


3. Calculate the pH of the equivalence point for titrating 0.200 M solutions of each of the following bases with 0.200 MHBr:
a. sodium hydroxide (NaOH)

4. Calculate the solubility of Mn(OH)2 in grams per liter when buffered at pH:
a. 7.0

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for chemistry ap experts,
i really need help with these problems...please help asap
thankssssssss

[joie.de.vivre]

1a. use the equation pH=pKa + log ([Concentration of Base]/[Concentration of acetic Acid])

1b. the KOH reacts with the acetate ion [from sodium acetate] to create acetic acid. the limiting reactant is KOH
use the equation again to find PH.

2a. find the moles for both HBr and NaOH to find the limiting reactant. then i think you should just use -log([concentration of H+ ion])

3a. the pH is 7

4. Ksp= solubilty= [Mn] x[OH]squared. let [Mn] be the variable x.
Ksp= x(2x)^2= 4x^3
you know that the concentration of OH- is 10^-7. plug that in for x. after you solve just use molar mass of Mn(OH)2 to get it into grams per liter.

asdfasdf i hope i'm not wrong on this; i learned this chapter a while ago @_@

[영원한 사랑]

^how do you calculate the ph of 7 though?
cuz i know the answer is 7 but i dont know how to get to it...thanks

[josebiwasabi]

^how do you calculate the ph of 7 though?
cuz i know the answer is 7 but i dont know how to get to it...thanks

equal concentration of a strong acid and strong base, i think.

[trixtah]

1b. the KOH reacts with the acetate ion [from sodium acetate] to create acetic acid. the limiting reactant is KOH
use the equation again to find PH.

I don't think I've ever seen KOH react with acetate ion to form acetic acid

KOH + C2H3O2- ~> HC2H3O2?

I'm pretty sure KOH reacts with acetic acid to form some acetate ion (KC2H3O2 dissociates)

er, HC2H3O2 + KOH ~> H2O + KC2H3O2

eww...seeing it without subscripts...shudder*