Question

A chemistry graduate student is given 125.mL of a 0.20M acetic acid HCH3CO2 solution. Acetic acid is a weak acid with =Ka×1.810−5. What mass of NaCH3CO2 should the student dissolve in the HCH3CO2 solution to turn it into a buffer with pH =4.47?

Answer 1

Answer : The mass of sodium acetate is, 1.097 grams.

Explanation : Given,

The dissociation constant for acetic acid =
`K_a=1.8* 10^(-5)`

Concentration of acetic acid (weak acid)= 0.20 M

volume of solution = 125. mL

pH = 4.47

First we have to calculate the value of
`pK_a`.

The expression used for the calculation of
`pK_a` is,

`pK_a=-\log (K_a)`

Now put the value of
`K_a` in this expression, we get:

`pK_a=-\log (1.8* 10^(-5))`

`pK_a=5-\log (1.8)`

`pK_a=4.74`

Now we have to calculate the concentration of sodium acetate (conjugate base or salt).

Using Henderson Hesselbach equation :

`pH=pK_a+\log ([Salt])/([Acid])`

Now put all the given values in this expression, we get:

`4.47=4.74+\log (([Salt])/(0.20))`

`[Salt]=0.107M`

Now we have to calculate the mass of sodium acetate.

`\text{Concentration}=\frac{\text{Mass of }NaCH_3CO_2* 1000}{\text{Molar mass of }NaCH_3CO_2* \text{Volume of solution (in mL)}}`

`0.107M=\frac{\text{Mass of }NaCH_3CO_2* 1000}{82g/mol* 125mL}`

`\text{Mass of }NaCH_3CO_2=1.097g`

Therefore, the mass of sodium acetate is, 1.097 grams.