Question

A buffer solution contains 0.496 M KHCO3 and 0.340 M K2CO3. If 0.0585 moles of potassium hydroxide are added to 250. mL of this buffer, what is the pH of the resulting solution

Answer 1

pH = 10.9

Step-by-step explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to say that the undergoing reaction between this buffer and OH⁻ promotes the formation of more CO₃²⁻ because it acts as the base, we can do the following:

`n_(CO_3^(2-))=0.34mol/L*0.250L+0.0585mol=0.1435mol\\\\n_(HCO_3^(-))=0.34mol/L*0.250L-0.0585mol=0.0265mol`

The resulting concentrations are:

`[CO_3^(2-)]=(0.1435mol)/(0.25L)=0.574M \\`

`[HCO_3^(-)]=(0.0265mol)/(0.25L)=0.106M`

Thus, since the pKa of this buffer system is 10.2, the change in the pH would be:

`pH=10.2+log((0.574M)/(0.106M) )\\\\pH=10.9`

Which makes sense since basic OH⁻ ions were added.

Regards!