Question

1.50 g of a weak acid (molar mass 176) is dissolved in 50.0 mL of water, and the resultant solution is titrated with 0.300 M NaOH. When 12.5 mL of 0.300 M NaOH is added, the pH of the resultant solution is 4.00. Calculate Ka for this acid.

Answer 1

The dissociation constant of this acid is
`7.85* 10^(-5)`.

Step-by-step explanation:

Moles of acid =
`(1.50 g)/(176 g/mol)=0.008523 mol`

Moles of sodium hydroxide:

`Moles=Concentration* volume (L)`

Moles of sodium hydroxide:

`=0.300 M* 0.0125 L=0.00375 mol`

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

`pH=pK_a+\log(([salt])/([acid]))`

`HA+NaOH\rightarrow NaA+H_2O`

Initial: 0.008523 0.00375 0

Final: (0.008523-0.00375) 0 0.00375

Volume of solution = 50.0 mL + 12.5 mL = 62.5 mL = 0.0625 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

`pH=pK_a+\log(([salt])/([acid]))`

`pH=pK_a+\log(([NaA])/([HA]))`

We are given:

`pK_a` = ?

`[HA]=(0.008523 mol-0.00375)/(0.0625 L)=(0.004773 mol)/(0.0625 L)`

`[NaA]=(0.00375 mol)/(0.0625 L)`

pH = 4.00

Putting values in above equation, we get:

`4.00=pK_a+\log (((0.00375 mol)/(0.0625 L))/((0.004773 mol)/(0.0625 L)))\\\\pK_a=4.105`

`4.105=-\log[K_a]`

`K_a=7.85* 10^(-5)`

The dissociation constant of this acid is
`7.85* 10^(-5)`.