Question

The Kb of weak base B is 1.0 x 10-6. A solution contains 0.75 M B and 0.25 M HB+Cl- (the salt of B with HCl). What is the pH of the solution after 0.05 mol NaOH is added to 1.0 L of the above solution?

Answer 1

Answer: pH of the given solution is 8.7.

Step-by-step explanation:

For the buffer mixture, initial pOH will be given as follows.

pOH =
`pK_(b) + log (salt)/(base)`

=
`-log(1 * 10^(-6)) + log (0.25)/(0.75)`

=
`6 - 0.477`

= 5.523

When 0.05 mol NaOH is added to this buffer solution then concentration of species present will be as follows.

`NaOH + HB^(+)Cl^(-) \rightarrow BOH + NaCl`

0.05 mol 0.25 0.75 0

0 0.20 (0.75 + 0.25) 0.05

Hence, the volume of solution will be 1 liter.

[BOH] =
`(1.00)/(1)`, [Salt] =
`(0.20)/(1)`

So, pOH =
`-log(1 * 10^(-6)) + log ((0.20)/(1))`[/tex]

= 6 - 0.69

= 5.30

Now, we will calculate the pH as follows.

pH = 14 - 5.30

= 8.7

Thus, we can conclude that pH of the given solution is 8.7.