If 25mL of 2.00M NAOH is added tpo 50mL of 0.10M HC2H3O2 (Ka=1.8*10^-5), what is the pH?
moles HC2H3O2= c*v = .1 * .05L = .005
moles NaOH= c*v = .2 * .025L = .005
pH= Pka + log (B/A)
= 4.75 + log (.005/.005)
= 4.75
I'm not sure how to calculate the moles I know i'm making a mistake there
Moles=concentration*volume
You got the moles wrong on NaOH
n = c* v
= .2mol/L * .025L = .005
is the c im using wrong or the V?
The original post says 25 mL of 2.00 M NaOH = 25 x 2 = 50 millimoles or 0.05 moles NaOH. But your solution says 0.2 x 0.025 = ??. Something is awry. Is it 2.00 M or 0.200 M? I think that's the trouble spot.
oh no that was a type it's actually .20M NaOH
so that gives me: n = c* v
= .2mol/L * .025L = .005
moles HC2H3O2= c*v = .1 * .05L = .005
I still get the same moles for both of them!
So, what do you have in the solution?
You have added 0.005 moles NaOH to 0.005 moles HC2H3O2. That produces 0.005 moles NaC2H3O2 (sodium acetate, the salt which is now dissolved in 75 mL water) + H2O from the neutralization reaction with no acid or base remaining unreacted. Your use of the H-H equation isn't justified BECAUSE you don't have a buffer solution. All you have is the salt of a weak acid and a strong base, namely NaC2H3O2.
The concn of the salt is 0.005 moles/0.075 mL. So it will hydrolyze as
(I like to call acetate Ac^----it sure saves a lot of typing).
Ac^- + HOH ==> HAc + OH^-
Kb = (Kw/Ka) = (OH^-)(HAc)/(Ac^-)
You know Kw, Ka, and (Ac^-). Solve for (OH^-) and pOH and pH from that.
To calculate the moles of a substance, you can use the formula: moles = concentration (in moles per liter) × volume (in liters).
For the acetic acid (HC2H3O2):
moles = concentration × volume
= 0.10 mol/L × 0.050 L
= 0.005 mol
For the sodium hydroxide (NaOH):
moles = concentration × volume
= 2.00 mol/L × 0.025 L
= 0.050 mol
So, the moles of HC2H3O2 is 0.005 mol, and the moles of NaOH is 0.050 mol.
Now, let's calculate the pH using the Henderson-Hasselbalch equation for acid-base reactions. The equation is:
pH = pKa + log(B/A)
where pKa is the negative logarithm of the acid dissociation constant and B/A is the ratio of the concentration of the base (NaOH) to the concentration of the acid (HC2H3O2).
Given that the Ka (acid dissociation constant) for acetic acid is 1.8 × 10^-5, the pKa is -log(1.8 × 10^-5) = 4.75.
Now, substituting the values into the equation:
pH = 4.75 + log (0.050/0.005)
= 4.75 + log (10)
= 4.75 + 1
= 5.75
Therefore, the pH of the solution after adding 25 mL of 2.00 M NaOH to 50 mL of 0.10 M HC2H3O2 is approximately 5.75.