Question

# Using 1.8x10-5 for the Ka for acetic acid, along with the weight of sodium acetate (4.0082g)...

Using 1.8x10-5 for the Ka for acetic acid, along with the weight of sodium acetate (4.0082g) and volume of acetic acid used to prepare the buffer (10.0 mL of 3.0M acetic acid), calculate the expected pH of the buffer. Show your work.

We created the buffer using 4.0082g sodium acetate, mixed with 10.0 mL of 3.0 M acetic acid and 90 mL of deionized water.

mass of sodium acetate=4.0082g

moles of sodium acetate=4.0082g/molar mass=4.0082/82.03g/mol=0.049 moles

concentration of sodium acetate=moles/volume of soution=0.049moles/(90+10)ml

=0.049mol/100ml *(10^3 L/ml)

=0.49 mol/L=0.49M

conc of acetic acid=3.0 M

ka=1.8*10^-5

pka=-logka=-log (1.8*10^-5)=5-0.255=4.74

pH=pKa +log [base]/[acid] {henderson-hasselbach equation}

or pH=4.74 +log[ acetate]/[acetic acid]

=4.74 + log (0.49/3.0)

=4.74 + log 0.163

=4.74-0.79

pH =3.75

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