Question

assume that for the pH 9.2 phenolphthalein sample, you used half as much phenolphthalein solution (i.e your sample is 9 ml buffer, 0.5 ml phenophthalein solution and 0.5 ml water). what should happen to the value of Kc you determine?

Answer 1

Step-by-step explanation:

It is given that pH is equal to 9.2. So, we will calculate the value of
`pK_(a)` as follows.

pH =
`pK_(a) + log \frac{[H_(3)O^(+)]}{[\text{solution}]}`

Putting the given values into the above formula as follows.

pH =
`pK_(a) + log \frac{[H_(3)O^(+)]}{[\text{solution}]}`

`pK_(a) = pH - log ((1)/(0.5))/((1)/(10))`

= 7.9

`K_(a) = 10^(7.9)`

=
`1.26 * 10^(-8)`

Since, the value of
`K_(a)` comes out to be positive. This means that there will occur an increase in the value of
`K_(a)` or
`K_(c)`.