acid base help?

Question

A 10.0-mL sample of 0.75 M CH3CH2COOH is titrated with 0.30 M NaOH. What is the pH of the solution after 22.0 mL of NaOH have been added to the acid? Ka = 1.3 × 10-5
5.75
4.94
4.83
4.02
3.95
plz help me, thanks in advance

Answer 1

definitely i dont know this one is buffer or not, who can guarantee the henderson equation is working or not? basicly you need to work wiff i diagram
CH3CH2COOH + NaOH-> Na+ +CH3CH2COO- +H20
B(bEfore) .0075mol
A(adding) ............... 0066mole
R(remain)....9x10^-4........... 0 .............. 0066mole
total volume is 32ml=.032L
after that react, remain is CH3CH2COOH and CH3CH2COO-
use i-diagram base on concetration
CH3CH2COOH+H2O->CH3CH2COO-+H30+
I .028125M.................................. 20625M....... 0
C -x................................................... X .............. X
E.0281-X........................................ 206+x.......... X
Ka=[CH3CH2COO-][H30+]/[CH3CH2COOH]
assume x is small by use 400 rule or 5% rule
Ka=x(.206)/.0281=1.3x10^-5
x= 1.77x10^-6=[H3O+]
ph=-log(1.77x10^-6)= 5.751=5.75 which is A.
good luck

Answer 2

10 ml of 0.75 M is 7.5 mmoles. 22 ml of 0.30 M is 6.6 mmol, so there will be 6.6 mmol of CH3CH2CO2- in solution and 0.9 mmol of CH3CH2CO2H in solution. Since both are in the same solution, the volume is irrelevant.

Use the Henderson-Hasselbach equation to figure out the pH.

pH = pKa + log (mmol base/mmol acid)
pH = 4.89 + log (6.6/0.9) = 4.89 + 0.87 = 5.75

maurizina would have been right, but didn't figure the concentration of the salt correctly. It would be 0.0066 moles/0.032 L or 0.206 M.

Answer 3

Moles of acid = 0.75(10)/1000=0.0075

Moles of OH- =0.30 (22)/1000 =0.0066

Total volume = 32 mL = 0.032 L

0.0075 mole acid + 0.0066 mole OH- react to give 0.0009 mole acid + 0.0066 mole salt

Concentration acid = 0.0009/0.032=0.0281 M

Concentration salt = 0.0066/0.0281= 0.235 M

pK = - log 1.3 10^-5 = 4.89

pH = 4.89 + log 0.235/0.0281=5.81