Question

How many moles of solid NaF would have to be added to 1.0 L of 0.115 M HF solution to achieve a buffer of pH 3.46? Assume there is no volume change. Ka for HF = 7.2 × 10 –4

Answer 1

Answer : The moles of solid NaF is, 1.09 mole.

Explanation : Given,

pH = 3.46

`K_a=7.2* 10^(-4)`

Concentration of HF = 0.115 M

Volume of solution = 1.0 L

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 (7.2* 10^(-4))`

`pK_a=4-\log (7.2)`

`pK_a=3.14`

Now we have to calculate the moles of HF.

`\text{Moles of }HF=\text{Concentration of }HF* \text{Volume of solution}`

`\text{Moles of }HF=0.115M* 1.0L=0.115mol`

Now we have to calculate the moles of NaF.

Using Henderson Hesselbach equation :

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

`pH=pK_a+\log \frac{\text{Moles of NaF}}{\text{Moles of HF}}`

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

`3.46=3.14+\log (\frac{\text{Moles of NaF}}{0.115})`

`\text{Moles of NaF}=1.09mol`

Thus, the moles of solid NaF is, 1.09 mole.